Understanding Elementary Shapes - Class 6 Maths - Extra Questions
Identify the triangle as equiangular, acute, obtuse or right.Answer: Obtuse
Mark answer as 1 if true else 0 if false
Mark answer as 1 if true else 0 if false
Identify the triangles as equilateral, isosceles or scalene.
Answer: :Isoceles
Mark answer as 1 if true else 0 if false
Mark answer as 1 if true else 0 if false
How many faces and edges does a triangular prism have?
If $$f$$, $$e$$ and $$v$$ represent the number of rectangular faces, number of edges and number of vertices respectively of a cuboid, then the values of $$f$$ is
Identify the triangles as equilateral, isosceles or scalene.
Answer: Scalene
Mark answer as 1 if true else 0 if false
Mark answer as 1 if true else 0 if false
Identify the triangle as equiangular, acute, obtuse or right.Answer: Acute
Mark answer as 1 if true else 0 if false
Mark answer as 1 if true else 0 if false
Identify the triangle as equiangular, acute, obtuse or right.
Answer: Right angled triangle
Mark answer as 1 if true else 0 if false
Mark answer as 1 if true else 0 if false
Define a Square.
Give a definition for Perpendicular lines
Read the following statement:
"A square is a polygon made up of four line segments, out of which, length of three line segments are equal to the length of fourth one and all its angles are right angles".
Define the terms used in this definition which you feel necessary. Are there any undefined terms in this? Can you justify that all angles and sides of a square are equal?
The terms used are:
1. Polygon: A simple closed figure made up of three or more line segments.
2. Line segment: Part of a line with two endpoints.
3. Line: Undefined term.
4. Point: Undefined term.
5. Angle: A figure formed by two rays with a common initial point.
6. Ray: Part of a line that has one endpoint and goes on infinitely in only one direction.
7. Right angle: Angle whose measure is $$90^\circ$$
1. Polygon: A simple closed figure made up of three or more line segments.
2. Line segment: Part of a line with two endpoints.
3. Line: Undefined term.
4. Point: Undefined term.
5. Angle: A figure formed by two rays with a common initial point.
6. Ray: Part of a line that has one endpoint and goes on infinitely in only one direction.
7. Right angle: Angle whose measure is $$90^\circ$$
There are two undefined terms in this which are line and point.
Consider a square $$ABCD$$
Given: The lengths of three line segments are equal to the length of the fourth one.
So,
$$AB = DA$$
$$BC = DA$$
and $$CD = DA$$
Given: The lengths of three line segments are equal to the length of the fourth one.
So,
$$AB = DA$$
$$BC = DA$$
and $$CD = DA$$
From Euclid's first axiom " Things which are equal to the same thing are also equal to one another", we get
$$AB = BC = CD = DA$$
And also given: All its angles are right angles.
$$AB = BC = CD = DA$$
And also given: All its angles are right angles.
From Euclid's fourth postulate "That all right angles are equal to one another", we get
$$\angle A = \angle B = \angle C = \angle D$$
$$\angle A = \angle B = \angle C = \angle D$$
So, we cansay that all angles and sides of a square are equal.
A man goes $$1 \cdot 2 km$$ due to North and then $$0 \cdot 5 km$$ due West. Find the least distance between his initial and final positions.
Based on the sides, classify the following triangles
Based on the sides, classify the following triangles
Based on the sides, determine the type of given triangle.
Based on the sides, classify the following triangles
Complete the following table
Fill up the blanks to make the statement true:A triangle cannot have more than _______________ obtuse angle.
Based on the sides, classify the following triangles
Based on the sides, determine the type of given triangle.
Complete the table by using Euler's formula.
| $$F$$ | $$8$$ | $$5$$ | ? |
| $$V$$ | $$6$$ | ? | $$12$$ |
| $$E$$ | ? | $$9$$ | $$30$$ |
What is the angle between two hands of a clock when the time in the clock is $$9'O$$ clock?
The time in the clock is $$9'O$$ clock.
We can clearly see from the figure, angle between the hands is $${ 90 }^{ \circ }$$ when the time is $$9'o$$ clock.Hence, the angle is $$90^o$$ or $$Right\ angle$$.
Using Euler's formula find $$V$$ if $$E=18$$, $$F=8$$.
Total number of faces in a cuboid is .................
A cuboid is made up of 6 rectangular faces.
Football is an example of __________ (Geometrical shape).
Shape of each face of Dodecahedron is ______
Look at the walls in your room. How many right angles are there? Count them.
Mark the right angles in the given pictures. You can mark more than one angle.
Make at least $$3$$ floor patterns using the given tile.
The sample 3 floor tiles pattern are given above . You can design it on your own too.
The first floor pattern is made using 18 tiles
The second floor pattern is made using 4 tiles
The third floor pattern is made using 10 tiles
Can you draw shapes on these dots? Make squares, rectangles and triangles of different sizes by joining the dots given. One has been done for you.
Square is formed by sides of equal lengths.
Triangles can be formed easily by joining 3 sides
Various rectangles , squares and triangles are drawn as shown in the figure.
Find the area of that triangle whose vertices are $$(-3,-2),(5,-2)$$ and $$(5,4)$$. Also prove that it is right angle triangle.
If $$m\angle XMQ = 90^{\circ}$$, then find the measure of remaining angles.
$$ABC$$ is a triangle, right angles at $$C$$. If $$AB = 25 cm$$ and $$AC = 7 cm$$, find $$BC$$.
ABC is a right triangle, right angled at C. If p is the length of perpendicular from C to AB & a, b, c have usual meaning, then prove that
(i) pc = ab
(ii) $$\frac{1}{{{p^2}}}\, = \,\,\frac{1}{{{a^2}}}\, + \,\frac{1}{{{b^2}}}$$
(i) Area of $$\Delta ABC = {1 \over 2}BC \times AC$$
$$ = {1 \over 2} \times a \times b = {{ab} \over 2}$$3
$$area = {1 \over 2} \times CD \times AB$$
$${1 \over 2} \times p \times c = {{pc} \over 2}$$
$${{ab} \over 2} = {{pc} \over 2}$$
$$ab = pc......(1)$$
(ii) In $$\Delta ABC$$
$$A{B^2} = A{C^2} + B{C^2}$$
$${\left( {{{ab} \over p}} \right)^2} = {b^2} + {a^2}$$
$${1 \over {{p^2}}} = {{{b^2} + {a^2}} \over {{a^2} + {b^2}}}$$
$${1 \over {{p^2}}} = {1 \over {{a^2}}} + {1 \over {{b^2}}}$$
$$\Delta$$ ABC is right angled at C. Then prove that $$\sin^2 A+\sin^2 B=1$$.
If $$8{R}^{2}={a}^{2}+{b}^{2}+{c}^{2}$$, then prove that the $$\triangle$$ is right angled.
Name the types of following triangles:
Triangle with length of sides $$7\ cm, 8\ cm $$ and $$9\ cm $$
The side opposite to the right angle in a triangle is called ................
What type of triangle is if one of the angle is $${90^ \circ }$$ or right angle ?
If one of the angles is $$90^o$$ then the triangle is called a right angled triangle.
Construct a triangle $$\Delta ABC$$ such that AB-AC =.3.5 cm, BC=8 cm,and $$\angle B=45$$.
$$(1).$$ Draw a line $$BC$$
$$(2).$$ Draw an arc of radius 8 cm. with $$B$$ as the centre.
$$(3).$$ $$AB=8 cm.$$
$$(4).$$ Draw an arc of radius $$8$$ $$cm.$$ with $$C$$ as the centre.
$$(5).$$ $$AC=8$$ $$cm.$$
$$(6).$$ Meet point $$A$$ to $$B$$ and $$C$$.
$$(7).$$ Then, Complete triangle $$ABC$$
In a right triangle ABC right angled at C, if $$\tan { A=1 } $$ then verify that 2$$\sin { A } \cos { A=1 } $$
If $$^ { \prime } A ^ { \prime }$$ is the area of a right triangle and $$^ { \prime } b ^ { \prime }$$ is the one of the sides containing the right angle, prove that lengths of the altitude on the hypotenuse from opposite vertex is $$\dfrac { 2 A b } { \sqrt { b ^ { 4 } + 4 A ^ { 2 } } }$$
Is there any possibility that a triangle has two right angles? Justify your answer.
No, a triangle cannot have two right angles because sum of three angles of a triangle is $$180^o$$. If two angles are of $$90^o$$ then one of the three angle has to be zero which is not possible.
I have no vertices.
I have no flat faces. Who am I?
Write all trigonometric ratio of the angle $$A$$ in right angle $$\Delta ABC$$
REF.Image
(i) Cos A = $$\frac{adjecent side}{hypotenuse}$$
cos A = $$\frac{AB}{AC}$$
(ii) Sin A = $$\frac{opposite side}{hypotenuse}$$
Sin A = $$\frac{BC}{AC}$$
(iii) tan A = $$\frac{opposite side}{adjecent side}$$
$$\Rightarrow $$ tan A $$= \frac{BC }{AB}$$
(iv) cot A = $$\frac{adjacent side}{opposite side}$$
cot A = $$\frac{AB}{BC}$$
(v) secent A = $$\frac{hypotenuse }{adjacent side}$$
secent A = $$\frac{AC}{AB}$$
(vi) cosec A = $$\frac{hypotenuse}{opposite side}\Rightarrow cosec A = \frac{AC}{BC}$$
In $$ \Delta A B C , A D \perp B C$$ such that $$A D ^ { 2 } = B D \times C D .$$ Prove that $$\Delta A B C$$ is right angle at $$A ?$$
$$\Delta $$ABC, AD $$\perp $$ CB, AC $$= 29$$ units, DC $$= 21$$ units and AB $$=40$$ units. Find the value of $$\cos^{2}B-\sin^{2}B$$
R.E.F image
Given $$ AD\perp BC$$ AC=29 unit
CD=21 unit AB=40unit
for, given figure
we have In $$\bigtriangleup ACD.$$
$$(29)^{2}=(21)^{2}+(AD)^{2}$$ [Pythagoras theorem]
$$(AD)^{2}=841-441=400$$
[AD=20] In $$\bigtriangleup ADB.$$
we know $$ (BD)^{2}=(AB)^{2}-(AD)^{2}=1600-400$$
$$(BD)^{2}$$ =1200
BD=20$$\sqrt{3}unit$$
now cos B =$$\frac{BD}{AB}$$ =$$\frac{20\sqrt{3}}{40}$$=$$\frac{\sqrt{3}}{2}$$ sin B
then $$cos^{2}B-sin^{2}B=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}$$
The angles of a triangle are in the ratio $$2:3:4$$. Determine the three angles. What type of triangle is it?
Angle of triangle are in ratio
2:3:4
Let angle of triangle are 2x,3x,&
4x then
2x+3x+4x=$$180^{\circ}$$ [angle sum property]
9x=$$180^{\circ}$$
$$[x=20^{\circ}]$$
then angle od trianle are =2x=$$40^{\circ}$$
$$=3x=60^{\circ}$$
$$=4x=80^{\circ}$$
This triangle is angled triangles.
The length of the sides of a right triangle are $$(2x-1)m, (4x)m$$ and $$(4x+1)m$$, where $$x>0$$. Find:
the value of $$x$$
the area of the triangle.
What type of a triangle is $$ABC$$ if
$$AB\neq BC\neq CA$$?
Explain how to construct $$150^{o}$$ using a pair of compasses.
What is scalene triangle ?
When all three sides of triangle are of different length, then triangle is scalene triangle.
The two sides of a right angled triangle are equal and the square of its hypotenuse is $$50\ \text{ cm }^{ 2 }$$. Find the length of each side.
In the given figure find $$OC$$
By Pythagoreans theorem
$$OA^2=OC^2+AC^2$$
$$\Rightarrow OC^2=\sqrt{OA^2-AC^2}$$.
In a $$\triangle ABC, D$$ is the mid point of side $$AC$$ such that $$BD =\dfrac{1}{2} AC$$. Show that $$\angle ABC$$ is a right angle.
Given that:
$$BD=\dfrac{1}{2}AC=AD=CD$$
$$\Rightarrow $$ Distance of $$D$$ is same from all vertices of $$\triangle ABC$$
Hence, $$D$$ is circumcentre of triangle ABC.
Now $$B$$ would be a point on circumcircle of $$\triangle ABC$$ and as angle in a semicircle is $$90^{\circ}$$, $$AC$$ being the diameter, we can say that:
$$\angle ABC=90^{\circ}$$
What is the measure of a right angle?
List ten situations where the angles made are right angles.
Take any two lines. Use the comparison by trace technique to compare the lengths.
To compare $$AB$$ and $$CD$$, we use a tracing paper, trace $$CD$$ and place the traced segment on $$AB$$ . Can you decide now which one among $$AB$$ and $$CD$$ is longer?
The method depends upon the accuracy in tracing the line segment. Moreover, if you want to compare with another length, you have to trace another line segment. This is difficult and you cannot trace the lengths everytime you want to compare them.
Draw a rough sketch of a quadrilateral KLMN. State two pairs of adjacent sides.
Two sides of a quadrilateral which have a common end point are called its adjacent sides.
For given diagram,
$$\overline{KL}, \overline{KN}$$ and $$\overline{NM}, \overline{ML}$$
$$\overline{KL}, \overline{LM}$$ and $$\overline{NM}, \overline{NK}$$.
$$\overline{KL}, \overline{LM}$$ and $$\overline{NM}, \overline{NK}$$.
From these two angles which has larger measure? Estimate and then confirm by measuring them.
Measure the angle given using the Protractor and write down the measure.
Measure the angle given using the Protractor and write down the measure.
Measure the angle given using the Protractor and write down the measure.
Measure the angle given using the Protractor and write down the measure.
Name the type of following triangle.
$$m\angle$$L$$=30^o$$, $$m\angle$$M$$=70^o$$ and $$m\angle$$N$$=80^o$$.
Name the following triangle in two different ways.(you may judge the nature of the angle by observation)
Name the following triangle in two different ways.(you may judge the nature of the angle by observation)
Name the type of following triangle.
Triangle with lengths of sides $$7\ \text{cm}$$, $$8\ \text{cm}$$ and $$9\ \text{cm}$$.
Name the types of given triangle:
$$\triangle DEF$$ with $$m\angle D = 90^{\circ}$$.
Name the type of the following triangle.
$$\Delta XYZ$$ with $$m\angle Y=90^o$$ and $$XY=YZ.$$
Name the type of following triangle.
$$\Delta$$ABC with AB$$=8.7$$cm, AC$$=7$$cm and BC$$=6$$cm.
Name the following triangle in two different ways.(you may judge the nature of the angle by observation)
Name the following triangle in two different ways. (you may judge the nature of the angle by observation)
If each of the two lines is perpendicular to the same line, what kind of lines are they to each other ?
If each of the following two lines is perpendicular to the same line, then they are parallel to each other. These lines will not intersect each other when extended infinitely in either of the two directions.
As given in the above figure,
$$m$$ and $$n$$ are $$\perp$$ to $$ l.$$
Also,
$$\angle 1 = \angle 2$$
which are corresponding angles so, $$m\parallel n.$$
Name the following triangle in two different ways.(you may judge the nature of the angle by observation)
What shape is a match box?
Step 1: To find
(1) Shape of a matchbox
Step 2: Shape of matchbox
Matchboxes are in general of "cuboid" shape.
Can a triangle have: Two obtuse angles?
Justify your answer in each case.
Calculate the size of angle BAC.
Angle BAC$$=$$ ________.
If one angle of a triangle is equal to the sum of the other two, show that the triangle is a right triangle.
In the given figure, verify by measurement that :
$$AB = 1.8\ cm, BC = 0.8\ cm, BD = 2.7\ cm, CD = 1.9\ cm, AC = 2.6\ cm$$ and $$AD = 4.5\ cm$$
$$AB + CD = AD - BC$$
In the given figure, compare the line segment with the help of a divider and fill in the blanks by using the symbol >, = or <
AB ..... CD
In the given figure, compare the line segment with the help of a divider and fill in the blanks by using the symbol >, = or <
BC ..... AB
In the given figure, compare the line segment with the help of a divider and fill in the blanks by using the symbol >, = or <
AC ..... BD
Name the following triangles with regards to sides :
Name the following triangles with regards to angles:
Classify the angles whose magnitudes are given below:
$$90^{o}$$
What is the shape of your geometry box?
What is the shape of a brick?
Name each of the following triangles in two different ways ( you may judge the nature of the angle by observation ):
Name each of the following triangles in two different ways ( you may judge the nature of the angle by observation ):
Name each of the following triangles in two different ways ( you may judge the nature of the angle by observation ):
Name each of the following triangles in two different ways ( you may judge the nature of the angle by observation ):
State which of the following statement are use true and which are false:
If all the three sides of a triangle are equal, it is called a scalene triangle.
State which of the following statement are use true and which are false:
An scalene triangle has all its angles equal.
A cuboid has ..... faces, ...... edges and ...... vertices.
Classify the angles whose measures are given below:
$$90^{o}$$
What is the shape of a drum ?
What is the shape of a match box ?
Fill in the blanks :
A cube has ..... square faces, .......... edges and ....... vertices.
What is the shape of a playing dice ?
Name the type of the following triangles:
(iv)$$\Delta XYZ$$ with $$\angle Y = 90^{o}$$ and XY = YZ.
Name the type of the following triangle:
(v)$$\Delta LMN $$ with $$\angle L = 30^{o}, \angle M = 70^{o}$$ and $$\angle N = 80^{o}$$
Name each of the following triangles in two different ways ( you may use ruler and protractor):
Name the type of the following triangles:
(iii)$$\Delta DEF$$ with D = $$\angle 90^{o}$$
Name each of the following triangles in two different ways ( you may use ruler and protractor):
Complete the table given
Name each of the following triangles in two different ways ( you may use ruler and protractor):
Count the number of cubes in the given shape
Look at the figure given below and classify the triangle according to its (a) sides (b) angles (You may judge the nature of the angle by observation):
Count the number of cubes in the given shape
How many faces, edges and vertices does a pyramid have with n sided polygon as its base?
Count the number of cubes in the given shape
In the given figures, identify the different shapes involved.
Look at the figures given below and classify each of the triangles according to its (a) sides (b) angles(You may judge the nature of the angle by observation):
Look at the figures given below and classify each of the triangles according to its (a) sides (b) angles(You may judge the nature of the angle by observation):
Look at the figures given below and classify each of the triangles according to its (a) sides (b) angles(You may judge the nature of the angle by observation):
Can the following be the sides of a right-angled triangle?
$$7cm, 5.6cm, 4.2cm$$
Look at the figures given below and classify each of the triangles according to its (a) sides (b) angles(You may judge the nature of the angle by observation):
State if the following can be the sides of a right angled triangle?
$$1.5cm, 2cm, 2.5cm$$
Classify the following triangles according to angle:
Classify the following triangles according to sides:
Look at the figures given below and classify each of the triangles according to its (a) sides (b) angles(You may judge the nature of the angle by observation):
Classify the following triangles according to sides:
The figure, given, shows shadow of some 3D object, when seen under the lamp of an overhead projector. Name the object.
The following figure is considered to be the image of a 3D object when seen under the lamp of an overhead projector
Classify the following triangles according to sides:
Match Column I with Column II
If $$c^{2}=a^{2}+b^{2}$$,then prove that $$4s\left (s-a \right )\left (s-b \right )\left (s-c \right )=a^{2}b^{2}$$.
Determine whether the triangle whose lengths of sides are 3 cm, 4 cm, and 5 cm is a right angled triangle?
The base of a right prism is a right angled triangle. The measure of the base of the right angled triangle is 3 m and its height 4 m. If the height of the prism is 7 m then find (i) the number of edges of the prism (ii) the volume of the prism (iii) the total surface area of the prism.
Let $$ABC$$ be an acute-angled triangle triangle, and let $$D,\,E,\,F$$ be points on $$BC,\,CA,\,AB$$ respectively such that $$AD$$ is the median, $$BE$$ is the internal angle bisector and $$CF$$ is the altitude. Suppose $$\angle FDE=\angle C,\,\angle DEF=\angle A\;and\;\angle EFD=\angle B$$. Prove that $$ABC$$ is equilateral.
Since $$\Delta BFC$$ is right-angled at $$F$$, we have $$FD=BD=CD=a/2$$. Hence $$\angle BFD=\angle B$$. Since $$\angle EFD=\angle B$$, we have $$\angle AFE=\pi-2\angle B$$. Since $$\angle DEF=\angle A$$, we also get $$\angle CED=\pi-2\angle B$$. Applying sine rule in $$\Delta DEF$$, we have
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\displaystyle\frac{DF}{sin A}=\displaystyle\frac{FE}{sin C}=\displaystyle\frac{DE}{sin B}$$.
Thus we get $$FE=c/2\;and\;DE=b/2$$. Sine rule in $$\Delta CED$$ gives
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\displaystyle\frac{DE}{sin C}=\displaystyle\frac{CD}{sin(\pi-2B)}$$.
Thus $$(b/sin C)=(a/2sin\,B\,cos\,B)$$. Solving for $$cos B$$, we have
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;cos B=\displaystyle\frac{a\,sin\,c}{2b\,sin \,B}=\displaystyle\frac{ac}{2b^2}$$.
Similarly, sine rule in $$\Delta AEF$$ gives
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\displaystyle\frac{EF}{sin\,A}=\displaystyle\frac{AE}{sin(\pi-2B)}$$.
This gives (since $$AE=b/c(a+c)$$), as earlier,
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;cos\,B=\displaystyle\frac{a}{a+c}$$.
Compairing the two values of $$cos\,B$$, we get $$2b^2=c(a+c)$$. We also have
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;c^2+a^2-b^2=2ca\,cos\,B=\displaystyle\frac{2a^2c}{a+c}$$.
Thus
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;4a^2c=(a+c)(2c^2+2a^2-2b^2)=(a+c)(2c^2+2a^2-c(a+c))$$.
This reduces to $$2a^3-3a^2c+c^3=0$$. Thus $$(a-c)^2(2a+c)=0$$. We conclude that $$a=c$$. Finally
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;2b^2=c(a+c)=2c^2$$
We thus get $$b=c$$ and hence $$a=c=b$$. This shows that $$\Delta ABC$$ is equilateral.
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\displaystyle\frac{DF}{sin A}=\displaystyle\frac{FE}{sin C}=\displaystyle\frac{DE}{sin B}$$.
Thus we get $$FE=c/2\;and\;DE=b/2$$. Sine rule in $$\Delta CED$$ gives
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\displaystyle\frac{DE}{sin C}=\displaystyle\frac{CD}{sin(\pi-2B)}$$.
Thus $$(b/sin C)=(a/2sin\,B\,cos\,B)$$. Solving for $$cos B$$, we have
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;cos B=\displaystyle\frac{a\,sin\,c}{2b\,sin \,B}=\displaystyle\frac{ac}{2b^2}$$.
Similarly, sine rule in $$\Delta AEF$$ gives
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\displaystyle\frac{EF}{sin\,A}=\displaystyle\frac{AE}{sin(\pi-2B)}$$.
This gives (since $$AE=b/c(a+c)$$), as earlier,
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;cos\,B=\displaystyle\frac{a}{a+c}$$.
Compairing the two values of $$cos\,B$$, we get $$2b^2=c(a+c)$$. We also have
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;c^2+a^2-b^2=2ca\,cos\,B=\displaystyle\frac{2a^2c}{a+c}$$.
Thus
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;4a^2c=(a+c)(2c^2+2a^2-2b^2)=(a+c)(2c^2+2a^2-c(a+c))$$.
This reduces to $$2a^3-3a^2c+c^3=0$$. Thus $$(a-c)^2(2a+c)=0$$. We conclude that $$a=c$$. Finally
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;2b^2=c(a+c)=2c^2$$
We thus get $$b=c$$ and hence $$a=c=b$$. This shows that $$\Delta ABC$$ is equilateral.
Write the following statement in five different ways conveying the same meaning
p : If triangle is equiangular then it is an obtuse angled triangle
Find the equation of the lines through the point $$(3, 2)$$ which make an angle of $$45\displaystyle ^{\circ}$$ with the line $$\displaystyle x-2y= 3$$.
In the figure, $$POQ$$ is a line. Ray $$OR$$ is perpendicular to line $$PQ$$. $$OS$$ is another ray lying between rays $$OP$$ and $$OR$$. Prove that
$$\angle ROS=\dfrac { 1 }{ 2 } \left( \angle QOS-\angle POS \right) $$.
The angles of a triangle are in AP and the number of degrees in the least is to the number of radians in the greatest as $$60 : \pi$$. Find the angles in degrees.
$$\Delta ABC$$ is an acute angled triangle CD be the altitude through C. If AB = 8, CD =Find the distance between the midpoints of AD and BC.
Let $$P$$ & $$R$$ are the mid-point of $$AD$$ & $$BC$$ respectively.
Now draw $$RQ$$ perpendicular to $$AB$$.
Now, in $$\triangle CDB $$ & $$ \triangle RQB,$$
$$\angle D=\angle Q=90°$$
$$\angle B=\angle B$$
$$\therefore \triangle CDB\sim\triangle RQB\text{ (by $$AA$$)}$$
So, basic propertionality theorem, if $$R$$ is mid point of $$BC$$.
$$\therefore Q$$ is midpoint of $$BD$$.
We can say,
$$\Longrightarrow AP+DQ=PD+QB$$
$$\Longrightarrow 2(AP+DQ)=PD+QB_AP+DQ$$
$$\Longrightarrow 2(AP+DQ)=AB$$
$$\Longrightarrow AP+DQ=\cfrac{AB}{2}$$
$$\Longrightarrow AP+DQ=\cfrac{8}{2}=4$$
Also,
$$QR=\cfrac{1}{2}CD=\cfrac{1}{2}\times 6=3$$
$$\therefore PR=\sqrt { { (QR) }^{ 2 }+{(PQ) } ^{ 2} }$$ (By pythagoras theorem)
$$=\sqrt { { (3) }^{ 2 }+{ (4) }^{ 2 } } $$
$$=\sqrt {25}$$
$$=5$$
In the given figure, $$\displaystyle \Delta ABC$$ is an obtuse triangle, obtuse angled at B. If $$\displaystyle AD\bot CB$$, then prove that $$\displaystyle { AC }^{ 2 }={ AB }^{ 2 }+{ BC }^{ 2 }+2BC.BD$$
Given: $$ABC$$ is obtuse triangle.
$${AB}^{2}={AD}^{2}+{DB}^{2}\longrightarrow (1)$$
$${AC}^{2}={AD}^{2}+{DC}^{2}\longrightarrow (2)$$
from $$(1)$$ $${AD}^{2}={AB}^{2}-{DB}^{2}$$
$$\therefore {AC}^{2}={AB}^{2}-{DB}^{2}+{(DB+BC)}^{2} [\because BC=DB+BC]$$
$${AC}^{2}={AB}^{2}-{DB}^{2}+{DB}^{2}+{BC}^{2}+2BC\times DB $$
$${AC}^{2}={AB}^{2}+{BC}^{2}+2BC\times DB $$ $$(Proved)$$
Which of the following will form the sides of a triangle(i) $$23cm,17cm,8cm$$ (ii) $$12cm,10cm,25cm$$, (iii) $$9cm,7cm,16cm$$
In the given figure, $$\angle X = 62^o, \angle XYZ = 54^o$$. In $$\triangle XYZ$$ If YO and ZO are the bisectors of $$\angle XYZ$$ and $$\angle XZY$$ respectively find $$\angle OZY$$ and $$\angle YOZ$$.
Show that the following points form a right angled triangle.
(2, -3), (-6, -7) and (-8, -3)
Show that the following points form a right angled triangle.
(10, 0), (18, 0) and (10, 15)
Show that the following points form a right angled triangle.
(5, 9), (5, 16) and (29, 9)
Show that the following points form an equilateral triangle.
$$(0, 0), (10, 0) \ and \ (5, 5 \sqrt 3)$$
Show that the following points form a right angled triangle.
(0, 0), (a, 0) and (0, b)
In the given figure $$\overleftrightarrow{PQ}$$ is a line. Ray $$\overrightarrow{OR}$$ is perpendicular to line $$\overleftrightarrow{PQ}. \overrightarrow{OS}$$ is another ray lying between rays $$\overrightarrow{OP}$$ and $$\overrightarrow{OR}$$.
Prove that $$\angle ROS = \dfrac{1}{2} (\angle QOS - \angle POS)$$
Construct $$\triangle DEF$$ with $$EF =4.5\ cm, \,DE =5\ cm$$ and $$DF=3\ cm$$
Given that
$$EF=4.5\ cm$$
$$DE=5\ cm$$
And,$$DF=5\ cm$$
step of construction
(i) Draw a line segment $$DE=5\ cm$$
(ii) With $$E$$ as centre and radius $$4.5\ cm$$ draw an arc.
(iii) With $$D$$ as centre and radius $$4\ cm$$ draw an arc cutting the previous arc at $$F$$
(iv) join $$EF$$ and $$DF$$ then $$\Delta DEF$$ is the required triangle
In $$\Delta$$ABC, BD$$\perp$$AC, D$$\in$$AC and $$\angle$$B is a right angle. If AC$$=5$$CD then show that BD$$=2$$CD.
From the triangle BDC.
$${ BC }^{ 2 }={ BD }^{ 2 }+{ x }^{ 2 }...(i)$$
and
From the triangle ABD
$${ AB }^{ 2 }={ BD }^{ 2 }+{ 16x }^{ 2 }...(ii)$$
From the triangle ABC
$${ AB }^{ 2 }+{ BC }^{ 2 }={ AC }^{ 2 }...(iii)$$
Thus substituting in (iii) from (i) and (ii),
$${ BD }^{ 2 }+{ x }^{ 2 }+{ BD }^{ 2 }+16{ x }^{ 2 }=(25{ x }^{ 2 })\\ 2{ BD }^{ 2 }=8{ x }^{ 2 }\\ { BD }^{ 2 }=4{ x }^{ 2 }\\ BD=2x$$
hence
$$BD=2CD$$
A triangle ABC is right angled at A. L is a point on BC such that AL $$\bot$$ BC. Prove that $$\angle$$ BAL =$$ \angle$$ ACB.
Since $$\Delta$$ ABC is right-angled,
$$AB^2+AC^2=BC^2$$
or, $$AB^2+AC^2=(BL+CL)^2$$
or, $$AB^2+AC^2=BL^2+CL^2+2.BL.CL$$.......(1).
Again from $$\Delta$$ ALC and $$\Delta$$ ALB we have
$$AC^2=AL^2+CL^2$$......(2) and $$AB^2=AL^2+BL^2$$.....(3).
Adding (2) and (3) we get,
$$AC^2+AB^2=2.AL^2+CL^2+BL^2$$
or, $$BL^2+CL^2+2.BL.CL=2.AL^2+BL^2+CL^2$$
or, $$AL^2=BL.CL$$.......(4).
Now $$\tan \angle ACB=\dfrac{AL}{CL}$$.......(5) and
$$\tan \angle ABL=\dfrac{AL}{BL}$$
$$\Rightarrow\tan (90^o-\angle BAL)=\dfrac{AL}{BL}$$
$$\Rightarrow\cot\angle BAL=\dfrac{AL}{BL}$$......(6).
From (4) we have $$\dfrac{AL}{CL}=\dfrac{BL}{AL}$$
or, $$\tan \angle ACB=\tan \angle BAL$$[Using (5) and (6)]
or, $$\angle ACB=\angle BAL$$.
$$ABC$$ is a right-angled triangle. Angle $$ABC = {90^o},$$ $$AC = 25\,cm$$ and $$AB = 24\,cm$$. Calculate the area of a $$\Delta ABC.$$
Using Pythagoras theorem,
$${\left( G \right)^2} = {\left( B \right)^2} + {\left( P \right)^2}$$
$${\left( {25} \right)^2} = {\left( B \right)^2} + {\left( {24} \right)^2}$$
$$625 = \left( {{B^2}} \right) + 576$$
$$625 - 576 = {B^2}$$
$$49 = {B^2}$$
$$\boxed{B = 7}$$
$$\therefore $$ Area of triangle $$ = \dfrac{1}{2} \times B \times h$$
$$ = \dfrac{1}{2} \times 7 \times 24$$
$$ = 84\,\,c{m^2}$$
$${\left( G \right)^2} = {\left( B \right)^2} + {\left( P \right)^2}$$
$${\left( {25} \right)^2} = {\left( B \right)^2} + {\left( {24} \right)^2}$$
$$625 = \left( {{B^2}} \right) + 576$$
$$625 - 576 = {B^2}$$
$$49 = {B^2}$$
$$\boxed{B = 7}$$
$$\therefore $$ Area of triangle $$ = \dfrac{1}{2} \times B \times h$$
$$ = \dfrac{1}{2} \times 7 \times 24$$
$$ = 84\,\,c{m^2}$$
The difference between the sides at right angles in a right angled triangle is 14 cm. The area of the triangle is 120 cm$$^2$$. Calculate the perimeter of the triangle.
Draw a right triangle in which the sides (other than hypotenuse) are of lengths $$5\ cm$$ and $$4\ cm$$. Then construct another triangle whose sides are $$\dfrac {5} {3}$$ times the corresponding sides of the given triangle.
Given , a right angled $$∆ ABC$$ , in which $$BC = 4 cm , AB= 3 cm$$ .
Here scale factor$$= \dfrac{5}{3} >1$$
Steps of construction:
1. Draw a line segment $$BC =4 cm.$$
2. At $$B$$ draw $$∠B = 90°$$ and cut off $$BA =3 cm$$ from it.
3. Join $$AC$$. Thus $$∆ABC$$ is a given right angled Triangle.
4. Now from $$B$$ draw a ray $$BY$$ making an acute $$∠CBY$$ on the side opposite to the vertex $$A$$.
5. Mark $$5$$ points $$B_1,B_2,B_3,B_4$$ & $$B_5$$ on $$BY$$ such that $$BB_1 = B_1B_2 = B_2B_3 = B_3B_4 =B_4B_5$$.
6.Join $$B_3C$$ and from $$B_5$$ draw $$B_5C’ || B_3C$$ intersecting $$BC$$ at $$C’$$.
7. From point $$C’$$ draw $$C’A’ || CA$$ intersecting $$AB$$ at $$A’$$.
Thus, $$∆A’BC’$$ is the required triangle whose sides are $$\dfrac {5}{3}$$ of the corresponding sides of $$∆ABC$$.
$$\angle B$$ is a right angle in $$\triangle ABC$$ and $$\bar{BD}$$ is an altitude to hypotenuse $$AB=8$$ cm and $$BC=6$$ cm. Find the area of $$\triangle BDC$$
The angle of elevation of top of a tower at a point on the ground is $$30^{\circ}$$. What is the angle of elevation if the height of tower is tripled?
given angle of elevation if height is h is 30 degrees from diagram we can say that
$$tan(30)=h/a$$
now if height is triples then $$tan(\theta)=3h/a$$
$$h=a/\sqrt{3}$$ implies
$$tan(\theta)=3h/a=3/\sqrt{3}=\sqrt{3}$$
therefore $$\theta =60$$
so the angle of elevation becomes $$60$$ degrees
There is a staircase as shown in the given figure, connecting point A, B. measurements of the step are marked in the figure. Find the straight line distance between A and B.
If you observe we can see that there is a large right triangle formed with the combined lengths of the other two sides of the triangle.
Extend the bottom-most triangle's base and topmost triangle's $$1.8\,unit$$ side till they touch each other at $$C$$.
$$\Rightarrow$$ $$AC=2+2+2+2=8\,unit$$
$$\Rightarrow$$ $$BC=1+1.6+1.6+1.8=6\,unit$$
$$\therefore$$ $$AB^2=AC^2+BC^2$$ [ By Pythagoras theorem ]
$$\Rightarrow$$ $$AB^2=(8)^2+(6)^2$$
$$\Rightarrow$$ $$AB^2=64+36$$
$$\Rightarrow$$ $$AB^2=100$$
$$\therefore$$ $$AB=10\,unit$$
$$\triangle$$MNP is right-angled triangle whose $$\angle MNP=90^\circ$$, $$ seg NQ\perp segMP$$, $$MQ=9$$, $$QP=4$$, find $$NQ$$.
$$b^{2}+ h^{2}= 13^{2}= 169$$..... $$(i)$$
also in $$\triangle QNP$$ :
$$QN^{2}+4 ^{2}= b^{2}$$...... $$(ii)$$
And in $$\triangle MNQ$$ :
$$QN^{2}+9^{2}= h^{2}$$....... $$(iii)$$
Adding $$(ii)$$ and $$(iii)$$;
$$QN^{2}+ 4^{2}+ QN^{2}+9^{2}= b^{2}+ h^{2}$$
$$\Rightarrow 2QN^{2}=13^{2}-4^{2}-9^{2}$$....... [From eq. $$(i)$$]
$$\Rightarrow 2QN^{2}=72$$
$$\Rightarrow QN^{2}=36$$
$$\Rightarrow QN=6\ cm$$
If a perpendicular is drawn from the vertex containing the right angle of a right triangle to the hypotenuse then prove that the triangle on each side of the perpendicular are similar is product of the lengths of the two parts of the hypotenuse.
$$\triangle ABC$$ right angled at $$B$$ and $$\bot$$ from $$B$$ intersecting $$AC$$ at $$D(BD\bot AC)$$
$$ In\triangle ADB\& \triangle ABC$$
$$ \angle A=\angle A\quad $$(common)
$$ \angle ADB=\angle ABC\quad $${ Each $${ 90 }^{ \circ }$$}
$$\triangle ADB\sim\triangle ABC$${ By AA Similarity criterion} -(1)
$$\\$$ Similarly,
$$ In\triangle BDC\& \triangle ABC$$
$$ \angle C=\angle C\quad $${ common}
$$ \angle BDC=\angle ABC\quad $${ Each $${ 90 }^{ \circ }$$}
$$ \triangle BDC\sim \triangle ABC\quad $${ By AA Similarity Criterion} -(2)
from$$(1)\& (2)$$
$$ \triangle ADB\sim \triangle ABC\quad \& \quad \triangle BDC\sim \triangle ABC$$
If one $$\triangle le$$ is similar to another $$\triangle le$$,and second$$ \triangle le$$ is similar, to the third $$\triangle le$$,then first & third $$\triangle le$$ are similar.
$$ \triangle ADB\sim \triangle BDC$$
If $$\Delta$$ ABC is a right-angled triangle prove that $$\sin^2 A + \sin^2 B + \sin^2 C = 2$$
If the sides of a triangle are in the ratio $$\,\,1:\,\sqrt 2 :1$$, show that it is a right-angled triangle.
If in a right angle triangle the hypotenuse is four times as long as the perpendicular Drawn to Opposite vertex, then one of its acute angle is
Ref Above figure
$$c^2 = h^2 + (a-x)^2$$
$$b^2 = h^2 + x^2$$
on adding, we have $$2h^2 + x^2 + (a-x)^2 = a^2$$ ..(1)
also $$a = 4h$$ and $$cos B = b/a = x/b$$
so $$x = b^2/a$$
on solving (1), we have $$h^2 + x^2 -4hx = 0$$
substituting above gives
$$b^4 + a^4/16 -a^2.b^2 =0 .. (2)$$
as $$cos B = b/a$$
on solving eqn. (2) we have , $$Cos^2B = 0.933$$ or $$0.067$$
$$Cos B = 0.966$$ or $$0.259$$
$$B = 15$$ or $$75$$ degrees
$$\triangle ABC$$ is an isosceles $$\triangle $$ with $$AB=AC$$ side $$BA$$ is produced to $$D$$ such that $$AB=AD$$ prove $$\angle BCD$$ is a right angle.
$$AB=AC$$
$$\Rightarrow \angle B=\angle ACB=\theta$$ (say)
Similarly $$AB=AD=AC$$
$$\Rightarrow \angle ACD=\angle ADC=\theta$$
$$\angle BAC={ 180 }^{ \circ }-2\theta $$(Consider the $$\triangle$$ ABC)
$$\angle CAD={ 180 }^{ \circ }-2\theta $$(Consider the $$\triangle$$ ADC)
Since $$BAD$$ is a straight line
$$\angle BAC+\angle CAD={ 180 }^{ \circ }\\ \Rightarrow { 360 }^{ \circ }-4\theta ={ 180 }^{ \circ }\\ \Rightarrow 4\theta ={ 180 }^{ \circ }\\ \Rightarrow \theta ={ 45 }^{ \circ }\\ \therefore \angle BCD=2\theta ={ 90 }^{ \circ }$$
In a right-angled triangle ABC, $$\angle B=90^o$$ then find the value of $$\sin A.\cos C+\cos A.\sin C$$.
The angles of some triangles are given below. Classify each triangle acute-angled,obtuse-angled or right-angled on the basis of its angles.
$$\begin{array}{l}a {.\ 90^\circ},{45^\circ},{45^\circ}\\b{.\ 60^\circ},{60^\circ},{60^\circ}\\c{.\ 80^\circ},{60^\circ},{40^\circ}\\d{.\ 92^\circ},{50^\circ},{38^\circ}\\e{.\ 120^\circ},{50^\circ},{10^\circ}\\f{.\ 90^\circ},{35^\circ},{55^\circ}\end{array}$$
Draw two angles of $${45^ \circ }$$ at the ends of a line segment. Extend these rays. What shape do yo get? Measure the angled formed.
Let $$AB$$ be the line segment
Draw $$45^o$$ at ends of line segment
Let they meet at $$C$$
$$\therefore $$ the figure $$ABC$$ is a triangle as it is a triangle the sum of angles $$=180^o$$
$$45+45+\angle C=180$$
$$\angle C=90^o$$
$$\therefore \ $$ the measure of the angle formed $$=90^o$$
A right angled triangle $$OPQ$$ is right angled at $$O$$. $$R$$ is the point on $$OQ$$
If $$PQ=17cm,QR=9cm$$ and $$PR=10cm$$. Find $$RO$$
In this figure there are two triangle $$\triangle POR\quad and\quad \triangle POQ$$ in both the triangle angle made at O is right angle .
Let the length of OR be x unit.
Apply pythagoras theorm for both the triangle
$$In t{ (PQ) }^{ 2 }={ (OQ) }^{ 2 }+({ OP })^{ 2 }\\ \Rightarrow { 17 }^{ 2 }={ (x+9) }^{ 2 }+{ (OP) }^{ 2 }\\ \Rightarrow { (OP) }^{ 2 }={ 17 }^{ 2 }-{ (x+9) }^{ 2 }$$$$\\ { (PR })^{ 2 }={ (OR) }^{ 2 }+{ (OP) }^{ 2 }\\ \Rightarrow { 10 }^{ 2 }={ x }^{ 2 }+{ (OP) }^{ 2 }\\ \Rightarrow { (OP) }^{ 2 }={ 10 }^{ 2 }-{ x }^{ 2 }$$
Since it is clear that the OP is same in both the equations ,so we can equate both the equations
$$\\ { 17 }^{ 2 }-{ (x+ }9)^{ 2 }={ 10 }^{ 2 }-{ x }^{ 2 }\\ \Rightarrow { 289 }-{ 100 }={ x }^{ 2 }+{ 18x }+{ 81 }-{ x }^{ 2 }\\ \Rightarrow { 189 }-{ 81 }={ 18x }\\ \Rightarrow { 18x }={ 108 }\Rightarrow { x }={ 6 }\ $$
What does below statement signifies.
$$AE\bot DC, AE\bot DC$$
Name the type of the triangle:
$$\triangle DEF$$ with $$\angle D=90^{o}$$
Without using the pythagoras theorem, Show that the point $$(4,4),(3,5) and (-1,-1)$$ are the vertices of a right angled triangle
In Figure, $$\angle\ X=62^{o}$$, $$\angle\ XYZ=54^{o}$$. If $$YO$$ and $$ZO$$ are the bisectors of $$\angle\ XYZ$$ and $$\angle\ XZY$$ respectively of $$\triangle\ XYZ$$, find $$\angle\ OZY$$ and $$\angle\ YOZ$$.
( Given) In $$\triangle xyz$$,
$$\angle x=62^o$$
$$\angle xyz =54^o$$
$$OY$$ is bisector of $$\angle xyz$$
$$OZ$$ is bisector of $$\angle xzy$$
To find $$\rightarrow \quad \angle ozy=?\quad \angle (yoz)=?$$
$$\Rightarrow $$ As we know, sum of angles of $$\triangle =180^o$$
$$\therefore \angle YXZ+\angle XYZ+\angle XZY=180^o$$
$$62+54+\angle XZY=180$$
$$\angle XZY =180-116$$
$$\angle XZY =64^o$$
As $$OZ$$ is bisector of $$\angle xzy$$
$$\therefore \angle OZY=\dfrac{\angle XZY}{2}=\dfrac{64^o}{2}$$
$$\boxed{\angle OZY =32^o}$$
$$\Rightarrow$$ Now, In $$\triangle OZY, \angle YOZ+\angle OYZ+\angle YZO=180^o$$
$$\angle YOZ+27+32=180^o$$
$$\angle YOZ=180-59$$
$$\boxed{\angle YOZ=21^o}$$
Name the types of following triangles :
$$\triangle LMN$$ with $$m\angle L = 30^{\circ}, m\angle M = 70^{\circ}$$ and $$m\angle N = 80^{\circ}$$.
If a $$\triangle ABC, \angle C$$ is an acute angle. $$AE\bot BC$$. Prove that $$AB^{2}=AD^{2}-BC\times DE+\dfrac {BC^{2}}{4}$$
We have,
$$ADE$$ is a right angled triangle.
Therefore ,By Pythagoras theorem,
$$ A{{E}^{2}}+E{{D}^{2}}=A{{D}^{2}} $$
$$ \Rightarrow A{{E}^{2}}=A{{D}^{2}}-E{{D}^{2}}\,\,......\,\left( 1 \right) $$
Again, $$In\,\Delta AEB$$
$$A{{B}^{2}}=A{{E}^{2}}+B{{E}^{2}}$$ ……. (2)
And now,
$$BE=BD-ED$$ …… (3)
Substituting equation (2) and (3) in (1), we get,
$$ A{{B}^{2}}=A{{D}^{2}}-E{{D}^{2}}+{{\left( BD-ED \right)}^{2}} $$
$$ =A{{D}^{2}}-E{{D}^{2}}+B{{D}^{2}}+E{{D}^{2}}-2BD\times ED $$
Now, $$BD=\dfrac{BC}{2}\,\,\,\,\,\,\,\,\,\,\,\left( \therefore \,D\,is\,\text{midpoint} \right)$$
Therefore,
$$ A{{B}^{2}}=A{{D}^{2}}+B{{D}^{2}}-2BD\times ED $$
$$ =A{{D}^{2}}+{{\left( \dfrac{BC}{2} \right)}^{2}}-2\left( \dfrac{BC}{2} \right)\times ED $$
$$ =A{{D}^{2}}-BC\times ED+\dfrac{B{{C}^{2}}}{4} $$
Hence, this is the answer.
In right triangle $$ABC$$, right angled at $$C$$, $$M$$ is the mid -point of hypotenuse $$AB$$. $$C$$ joined to $$M$$ and produced to a point $$D$$ such that $$DM = CM$$. Point $$D$$ is joined to point $$B$$ . Show that:
(i) $$\triangle AMC \cong BMD$$
(ii) $$\angle DBC $$ is a right angle.
(iii) $$\triangle DBC \cong \triangle ACB$$
(iv) $$CM =\dfrac{1}{2} AB$$
In $$\triangle ABC, \angle C=90^o$$
$$\to $$ In $$\triangle DMB$$ & $$\triangle AMC$$
$$DM=MC$$ (Given)
$$AM=MC$$ (Mid point of $$AB$$ is $$M$$)
$$\angle DMB= \angle AMC=(VOA)$$ (Velocity opposite angle)
$$\therefore $$\triangle DMB\cong \triangle AMC$$ (Anu $$SAS$$ rule
then,
$$DB=AC$$ (By $$CPCT$$)
As $$M$$ is the mid point of $$AB$$, so $$MC$$ bisect $$\angle C$$ in the $$\triangle ABC$$,
$$\angle MCB=\angle MCA=45^o$$
$$\angle MCA=\angle MOB$$ (Alternate angles)
$$\angle MDB=45^o=\angle CDB$$
& $$\angle MCB=\angle DCB=45^o$$
in $$\triangle DBC$$,
$$\angle CDB+\angle DBC +\angle DCB=180^o$$ [sum of angle in $$\triangle =180$$]
$$45^o+45^o+\angle DBC=180^o$$
$$\angle DBC=90^o$$
$$\therefore \ \angle DBC$$ is a right angle
In $$\triangle DBC$$ & $$\triangle ACB$$,
$$BC=BC$$ (common )
$$\angle ACB=\angle DBC=90^o$$
$$DB=AC$$ (proved through $$CPCT$$)
$$\therefore \ \triangle DBC\cong \triangle ACB$$ (By $$SAS$$ rule)
$$AB=CD $$ (By $$CPCT$$)
$$MC=\dfrac {1}{2} (CD)\quad $$ (as $$DM=MC$$)
$$=\dfrac {1}{2}(AB)\quad $$ (taken from put $$(w)$$)
Hence proved.
Fill in the blanks.
A solid cuboid has ____dimensions ?
In figure, $$PQR$$ is a triangle, right angled at $$Q$$. $$X$$ and $$Y$$ are the points on $$PQ$$ and $$QR$$ such that $$PX:XQ=1:2$$ and $$QY:YR=2:1$$. Prove that $${9({PY}^{2}+{XR}^{2})={13PR}^{2}}$$.
In the given figure $$\angle ABC = 90^\circ $$. And $$BD \bot AC$$. If $$AB=5.7 cm, BD=3.8 cm$$, and $$CD=4.4 cm$$, find length of side $$BC$$.
Given that,
$$\angle ABC={{90}^{0}}\,\And \,\,BD\bot AC$$
and
$$ AB=5.7\,cm $$
$$ BD=3.8cm $$
$$ CD=4.4cm $$
$$BC=?$$
Let $$AD=x\,cm$$ and $$BC=y\,cm$$
Then,$$In\,\Delta ABD$$
By pythagoras theorem,
$$ A{{B}^{2}}=A{{D}^{2}}+D{{B}^{2}} $$
$$ {{\left( 5.7 \right)}^{2}}={{y}^{2}}+{{\left( 3.8 \right)}^{2}} $$
$$ 32.49={{y}^{2}}+13.3 $$
$$ {{y}^{2}}=32.49-13.3 $$
$$ {{y}^{2}}=19.19 $$
$$ y=4.38\,cm $$
Now, $$In\,\Delta ABC$$
By Pythagoras theorem,
$$ A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}} $$
$$ {{\left( y+4.4 \right)}^{2}}={{\left( 5.7 \right)}^{2}}+{{x}^{2}} $$
$$ {{\left( 4.38+4.4 \right)}^{2}}=32.49+{{x}^{2}} $$
$$ {{\left( 8.78 \right)}^{2}}-32.49={{x}^{2}} $$
$$ {{x}^{2}}=77.87-32.49 $$
$$ {{x}^{2}}=45.38 $$
$$ x=6.73\,cm $$
$$ BC=x=6.73\,cm $$Hence, this is the answer.
In a triangle $$ABC$$, which is the longest side if :
(i) $$ \angle A\ $$ is a right angle
(ii) $$ \angle C\ $$ is a right angle
In right angle triangle $$\Delta P Q R ,$$ right angle is at $$Q$$ and $$P Q = 6 c m \, , \angle R P Q = 60 ^ { \circ } .$$ Determine the lengths of $$QR$$ and $$PR.$$
$$\\From \, the \, above \, figure$$
$$PR \cos60^0=6$$
$$\therefore\>PR=(\dfrac{6}{cos60^0})=12 $$
$$QR=PR \sin60^0$$
$$=12\times(\dfrac{\sqrt{3}}{2})=6\sqrt{3}$$
The area of right triangular region is $$129.5\ cm^{2}$$. If one of the sides containing the right angle is 14.8, find the other one.
In the figure given alongside, prove that
$$( i ) A B = F C$$
$$( i i ) \quad A F = B C$$
$$[Hint.$$ $$( i ) \Delta A B E \cong \Delta C F D , R H S$$
$$(ii) A B = F C \Rightarrow A F + F B = F B + B C .]$$
Given
9n triangle $$\Delta ABE$$ and $$\Delta CFD$$
(i) $$\angle ABE = \angle CFD $$ (RIght angled)
$$BE = FD$$ {GIven}
$$AE = DC$$
$$\therefore$$ from RHS rule
$$\Delta ABE \cong \Delta CFD$$
$$\therefore AB = FC$$ (from congruency Rule)
(ii) 9n the above part we have proved that
$$AB = FC \rightarrow AB = AF + FB$$
$$FC = FB + BC$$
$$\therefore AF + FB = FB + BC$$
$$\therefore AF = BC$$ Hence proved
In $$\Delta$$ABC, $$\angle$$A is obtuse, PB$$\perp$$AC and QC$$\perp$$AB. Prove that $$BC^2=(AC\times CP+AB\times BQ)$$.
Given $$\angle P=\angle Q=90^\circ$$
In $$\triangle ABP$$
$${AB}^{2}={AP}^{2}+{BP}^{2} $$(pythagoras theorem)
In $$\triangle BPC$$
$${BC}^{2}={BP}^{2}+{PC}^{2} $$(pythagoras theorem)
$${BC}^{2}={AB}^{2}-{AP}^{2}+{(AP+AC)}^{2}$$
$${BC}^{2}={AB}^{2}+{AC}^{2}+2AP\times AC.....(2)$$
In $$\triangle AQC$$
$${AC}^{2}={AQ}^{2}+{QC}^{2} $$(pythagoras theorem)
In $$\triangle BQC$$
$${BC}^{2}={CQ}^{2}+{BQ}^{2}$$ (pythagoras theorem)
$${BC}^{2}={AC}^{2}-{AQ}^{2}+{(AB+AQ)}^{2}$$
$${BC}^{2}={AC}^{2}+{AB}^{2}+2ABAQ......(3)$$
add (2) and (3)
$$2{BC}^{2}=2{AC}^{2}+2{AB}^{2}+2AP\times AC+2AB\times AQ$$
$$2{BC}^{2}=2AC(AC+AP)+AB(AB+AQ)$$
$$2{BC}^{2}=2(AC\times PC)+2(AB\times BQ)$$
$${BC}^{2}=AC\times CP+AB\times BQ$$
Hence proved
Find the measure of the supplement of $$(i){70}^{\circ}\,\,(ii){20}^{\circ}\,\,(iii){135}^{\circ}\,\,(iv){90}^{\circ}$$
The sides $$AB$$ and $$AC$$ of a triangle $$ABC$$ are produced; and the bisectors of the external angles at $$B$$ and $$C$$ meet at $$P$$. Prove that if $$AB > AC$$, then $$PC > PB$$.
Find the values of $$x$$ and $$y$$ in the figure given below.
Two acute angles of a right triangles are equal. Find the two angles.
Let the two equal acute angle be $$x$$ then, sum of all angles of a triangle
$$\therefore \ x+x+90^o=180^o$$
$$\Rightarrow \ 2x+90^o=180^o$$
$$\Rightarrow \ 2x=180^o-90^o$$
$$\Rightarrow \ 2x=90^o$$
$$\Rightarrow \ x=\dfrac {90^o}{2}$$
$$\Rightarrow \ x=45^o$$
$$\therefore \ $$ Two acute $$\angle s$$ are $$45^o, 45^o$$
In the adjoining figure, $$\triangle ABC$$ is right-angled at $$B$$ and $$\angle A = {45}^{\circ}$$. If $$AC = 3\sqrt {2}cm$$,
Find (i) $$BC$$,(ii) $$AB$$.
From right angled $$\triangle ABC$$,
(i) $$\dfrac {BC}{AC} = \sin 45^{\circ}$$
$$\dfrac {BC}{3\sqrt {2}} = \dfrac {1}{\sqrt {2}}$$
Or $$BC = 3$$
(ii) By Pythagoras theorem
$$(AB)^{2} = (AC)^{2} - (BC)^{2}$$
$$= (3\sqrt {2})^{2} - (3)^{2}$$
$$= 18 - 9$$
$$= 9$$
$$AB = 3\ cm$$.
(i) $$\dfrac {BC}{AC} = \sin 45^{\circ}$$
$$\dfrac {BC}{3\sqrt {2}} = \dfrac {1}{\sqrt {2}}$$
Or $$BC = 3$$
(ii) By Pythagoras theorem
$$(AB)^{2} = (AC)^{2} - (BC)^{2}$$
$$= (3\sqrt {2})^{2} - (3)^{2}$$
$$= 18 - 9$$
$$= 9$$
$$AB = 3\ cm$$.
In the adjoining figure, $$PQR$$ is a right triangle, right angled at $$Q,\ X$$ and $$Y$$ are the points on $$PQ$$ and $$QR$$ such that $$PX:XQ=1:2$$ and $$QY:YR=2:1$$. Prove that $$9(PY^{2}+XR^{2})=13PR^{2}$$.
The hypotenuse of a right-triangle is $$3\sqrt {10}\ \text{cm}$$. If the smaller side is tripled and the longer side is doubled, the new hypotenuse will be $$9\sqrt {5}\ \text{cm}$$. Find the length of each side.
Verify that points $$P(-2,2),Q(2,2)$$ and $$R(2,7)$$ are vertices of a right angled triangle.
R.E.F image
using distance formula,
$$AB=\sqrt{(-2-2)^{2}}+(2+7)^{2}$$
$$=\sqrt{16+25}$$
$$=\sqrt{41}$$
$$BC=\sqrt{(2+2)^{2}+(2-2)^{2}}$$
$$=\sqrt{4^{2}+0^{2}}$$
=4
$$AC=\sqrt{(2-2)^{2}+(2-7)^{2}}$$
$$=\sqrt{25}$$
=5
$$AC^{2}+BC^{2}=25+16=41=AB^{2}$$
$$\therefore$$ It is a right $$\angle 'ed \bigtriangleup $$
A solid has how many dimensions ?
Is the triangle with sides $$12cm$$, and $$18\ cm$$ a right angled triangle? Give reason.
'
If the bisector of the base angles of a triangle enclosed an angle of $$135^{o}$$, prove that the triangle is a right triangle.
Three angles of a triangle are $${ x }^{ \circ }$$, $$2{ x }^{ \circ }$$ and $$3{ x }^{ \circ }$$. Prove that the triangle is right angled.
Prove that the bisectors of the angles of a linear pair are at right angle.
Given:-
$$\angle{ACD}$$ and $$\angle{BCD}$$ are linear pairs.
$$CE$$ and $$CF$$ bisect $$\angle{ACD}$$ and $$\angle{BCD}$$ respectively.
To prove: $$\angle{ECF} = 90°$$
Proof:-
As we know that angle on a straight line is $$180°$$.
$$\therefore ∠ACD + ∠BCD = 180°$$
$$\Rightarrow 2 \angle{ECD} + 2 \angle{DCF} = 180° \quad \left( \because \text{CE and CF bisects } \angle{ACD} \text{ and } \angle{BCD} \text{ respectively} \right)$$
$$\Rightarrow \angle{ECD} + \angle{DCF} = \cfrac{180°}{2} = 90°$$
$$\therefore \angle{ECF} = 90°$$
Hence Proved.
In a $$\Delta$$ABC, right angled at B, $$AB = 24$$ cm, $$BC = 7$$ cm. Determine
(i) $$\sin A, \cos A$$
(ii) $$\sin C, \cos C$$
R.E.F image
In $$\Delta $$ ABC, B is at right angle.
Given, AB=24cm
BC=7cm
using Pythagoras theorem
$$AB^{2}+BC^{2}=AC^{2}$$
$$\Rightarrow (24)^{2}+(7)^{2}=AC^{2}$$
$$\Rightarrow AC=\sqrt{(24^{2})+(7)^{2}}=\sqrt{576+49}=\sqrt{625}$$
$$\Rightarrow AC=25 CM$$
(i)sin A=$$\frac{BC}{AC}$$
$$ = \frac{7}{25}$$
cos A=$$\frac{AB}{AC}$$
$$=\frac{24}{25}$$
(ii)sin C=$$\frac{AB}{AC}$$
$$=\frac{24}{25}$$
cos c=$$\frac{BC}{AC}$$
=$$\frac{7}{25}$$
Let $$A ( h , k ) , B ( 1,1 )$$ and $$C ( 2,1 )$$ be the vertices of a right angled triangle with $$A C$$ as its hypotenuse. If the area of the triangle is $$1$$ sq units, then the set of values which $$k$$ can take is given by
Let A, B, C are the vertices of a right angled triangle with AC as its hypotenuse.
Using Pythagoras theorem we get
$$AC^2=AB^2+BC^2$$
$$(h-2)^2+(k-1)^2=(h-1)^2+(k-1)^2+1$$
$$h^2+4-4h=h^2+1-2h+1$$
$$2h=2$$
$$[h=1]$$
Now area of triangle $$=\dfrac{1}{2}\times AB\times BC$$
$$1=\dfrac{1}{2}\times 1\times \sqrt{(h-1)^2+(k-1)^2}$$
$$(h-1)^2+(k-1)^2=4$$
$$h=1$$
$$(k-1)^2=4$$
$$k-1=\pm 2$$
$$k=3, -1$$
Hence k can take values $$(-1, 3)$$.
In $$\triangle {DFH}$$, point $$E$$ is the midpoint of hypotenuse $$DF$$ segment $$HE$$ is perpendicular to the diagonal $$DF$$, segment $$EG$$ is perpendicular to segment $$FH$$,segment $$EK$$ is perpendicular to segment $$DH$$.Show that $${EG}^{2}=FG\times EK$$
REF.Image
Given : $$\angle G=90^{\circ}=\angle K=\angle H=\angle DEH$$
$$\therefore EGHK$$ is a square
EF=ED
To prove : $$EG^{2}= FG\times EK$$
in $$\Delta $$ EFH & EHD
EF=ED (Given) _______ (1)
EH $$\Rightarrow $$ common sie ________ (2)
$$\angle DEH=\angle FEH=90^{\circ}$$
$$\therefore \Delta DEH\cong \Delta FEH$$ (SAS congruency)
$$\Rightarrow FH= DH$$
$$\Rightarrow \angle DFG=\angle FDG= 45^{\circ}$$
In $$\Delta FGE$$
tan < EFG = $$\frac{EG}{FG}$$
$$\Rightarrow tan 45 = \frac{EG}{FG}$$
$$\Rightarrow EG=FG$$ _______ (3)
EG=EK [ EGHK is square]________ (4)
from (3) & (4)
$$(EG)^{2}= FG\times EK$$
In Fig $$14.5$$, $$BD$$ is the median in the right isosceles $$\triangle ABC$$ $$BS$$ is equal to $$AB$$
$$\angle BAC=\angle BCD$$ & $$AB$$.$$BC$$ (Isoceles rt. angle triangle)
Hence $$\Delta ABD$$ & $$\Delta BDC$$ are congruent by SAS.
So $$\angle ADB=\angle BDC=\dfrac { 180 }{ 2 } ={ 90 }^{ 0 }$$
In $$\Delta ADB$$
$$AD$$.$$BD$$ Hence proved.
Find $$HCF$$ for the following set of number $$867$$ & $$255$$
Copy and complete the table for a triangle.
| Base | $$6$$ | $$8$$ | $$5$$ | $$x$$ | $$16$$ | ? | ? |
| Height | $$10$$ | $$14$$ | $$20$$ | $$2x$$ | ? | $$5$$ | $$11$$ |
| Area | ? | ? | ? | ? | $$120$$ | $$50$$ | $$110$$ |
Area $$=\dfrac { 1 }{ 2 } \times $$ base $$\times $$ height.
In figuer, $$BD$$ is the median in the right isosceles $$\Delta ABC,\ BD$$ is equal to $$AB$$
Name the following triangle :
$$\angle A=60^{o}, \angle B=90^{o}, \angle C=30^{o}$$
Since $$\angle B = 90^{\circ}$$ ,Therefore it is an right angle triangle right angled at B.
In $$\Delta ABC,\ m\angle B < 90$$ and $$\overline {AD}\bot \overline {BC}$$. Also $$AB=9,\ BC=6$$ and $$BD=2$$ then find $$AC$$.
By Pythagoras theorem in $$\Delta ADB$$,
$$AB^2=BD^2+AD^2$$
$$\Rightarrow AD^2=AB^20-BD^2$$
Similarly in $$\Delta ACD, AC^2=AD^2+DC^2$$
$$\Rightarrow AC^2=AB^2-BD^2+DC^2$$
$$=81-4+16=93$$
$$\Rightarrow AC=\sqrt{93}cm$$.
Draw the following angle using ruler and compasses. Also label them.
$$120^{o}$$
An angle of $${120}^{\circ}$$ is nothing but twice of an angle of $${60}^{\circ}$$.Therefore, it can be constructed as follows :
Step $$1:$$ Draw any line $$PQ$$ and take a point $$O$$ on it.
Step $$2:$$ Place the pointer of the compasses at $$O$$ and draw an arc of convenient radius which cuts the line at $$A$$.
Step $$3:$$ Without disturbing the radius on the compasses, draw an arc with $$A$$ as centre which cuts the first arc at $$B$$.
Step $$4:$$ Again without disturbing the radius on the compasses and with $$B$$ as centre, draw an arc which cuts the first arc at $$C$$.
Step $$5:$$ Join $$OC$$, $$\angle{COA}$$ is the required angle whose measure is $${120}^{\circ}$$.
In the given figure, if $$\angle ACB= \angle CDA, AC=8\ cm$$ & $$AD=3\ cm$$, then find $$BD$$.
In each of the following case, construct a point that is symmetric to the given point P with respect to the given line AB.
$$ABC$$ is a right triangle in which $$\angle B={ 90 }^{ \circ }$$. A circle is inscribed in the triangle. If $$AB = 8$$ cm and $$BC = 6$$ cm. find the radius $$r$$ of the circle.
In a right angle triangle, two acute angles are in the ratio, $$2:3$$. Find the angles.
In $$\triangle\ PQR$$ right angled at $$Q,\ PR+QR=25\ cm$$ and $$PQ=5\ cm$$. Determine the values of $$\sin P,\ \cos P$$ and $$\tan P$$.
In the adjoining figure, the area of the right angled triangle is $$54\;\text{cm }^{2}$$. If one of its legs is $$12\;\text{cm}$$ long, find its perimeter.
In the given diagram,
Area of right-angled triangle $$ =54 \;\text{cm}^{2} $$
Length of one leg $$ \mathrm{AB}=12 \;\text{cm} $$
Length of one leg $$ \mathrm{AB}=12 \;\text{cm} $$
Area = $$\dfrac{1}{2}\times $$ base $$\times $$ height
$$54=\dfrac{1}{2} \times $$ base $$\times 12$$
$$54=\dfrac{1}{2} \times $$ base $$\times 12$$
base $$=9\;\text{cm}$$
Now,
$$ {AC}^{2}={AB}^{2}+{BC}^{2} $$ (Pythagoras Theorem)
$$\begin{array}{l}=12^{2}+9^{2}=144+81=225=(15)^{2} \\A C=15 \mathrm{cm} \\\text { Perimeter }={AB}+{BC}+{AC}=12+9+15=36 \mathrm{cm}\end{array}$$
$$\begin{array}{l}=12^{2}+9^{2}=144+81=225=(15)^{2} \\A C=15 \mathrm{cm} \\\text { Perimeter }={AB}+{BC}+{AC}=12+9+15=36 \mathrm{cm}\end{array}$$
The side of a right triangle containing right angle are $$5\ cm$$ and $$(3x-1)cm$$.If area of tiangle is $$60\ cm^ {2}$$, find sides of triangle.
A $$15$$m long ladder reaches a window $$12$$m high from the ground on placing it against a well. How far is the ladder from the well?
R.E.F image
By using Pythagoras theorem,
$$(15)^{2}=(12)^{2}+(BC)^{2}$$
$$\Rightarrow 225= 144+(BC)^{2}$$
$$\Rightarrow (BC)^{2}=225-144$$
$$\Rightarrow (BC)^{2}=81$$
$$\Rightarrow BC=\sqrt{81}$$
$$\Rightarrow BC=9cm$$
BC = distance =9cm
wall is for away from the ladder by 9 meter.
$$ABC$$ is a right triangle right angled at $$C.$$ Let $$BC=a, CA=b,AB=c$$ and let $$p$$ be the length of perpendicular from $$C$$ on $$AB$$ prove that i) $$pc=ab$$ii) $$\dfrac{1}{{{p^2}}} = \dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}$$
In $$\triangle ABC,$$ $$CD$$ perpendicular to $$AB.$$
(i) To Prove:- $$pc=ab$$
In $$\triangle ABC,$$
Considering base $$BC$$ and height $$AC,$$
$$ar(ABC)=\dfrac 12 \times a\times b\quad\quad\quad\dots(i)$$
Considering base $$AB$$ and height $$CD,$$
$$ar(ABC)=\dfrac 12 \times p\times c\quad\quad\quad\dots(ii)$$
From equation $$(i)$$ and $$(ii),$$
$$\begin{aligned}{}\frac{1}{2} \times p \times c &= \frac{1}{2} \times a \times b\\pc& = ab\end{aligned}$$
(ii) Now,
$$\begin{aligned}{}pc& = ab\\\frac{c}{{ab}} &= \frac{1}{p}\\\frac{{{c^2}}}{{{a^2}{b^2}}} &= \frac{1}{{{p^2}}}\\\frac{{{a^2} + {b^2}}}{{{a^2}{b^2}}} &= \frac{1}{{{p^2}}}\quad\quad\quad[\text{By using Pythagoras Theorem}]\\\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} &= \frac{1}{{{p^2}}}\end{aligned}$$
Hence, proved.
Can a triangle have two obtuse angles? Give reason for your answer.
Show that in a right angled triangle, the hypotenuse is the longest side.
In $$\Delta ABC,\angle ABC = 135^\circ $$
Prove that:
$$A{C^2}$$=$$A{B^2}$$+ $$B{C^2}$$ + $$4(\Delta ABC)$$
Prove that:
$$A{C^2}$$=$$A{B^2}$$+ $$B{C^2}$$ + $$4(\Delta ABC)$$
Lets draw altitude from A to BC such that it meets BC at D, angles are $$\Delta ADC, \angle D=90^o$$
$$\Rightarrow AD^2+DC^2=AC^2$$
$$\Rightarrow AD^2+(BD+BC)^2=AC^2$$
$$\Rightarrow AD^2+BD^2+BC^2+2BD.BC=AC^2$$ ………….$$(1)$$
In $$\Delta ADB, \angle D=90^o$$
$$\Rightarrow AD^2+BD^2=AB^2$$
Substituting in equation $$(1)$$
$$\Rightarrow AD^2+BD^2+BC^2+2BD.BC=AC^2$$
$$\Rightarrow AB^2+BC^2+2BD.BC=AC^2$$
as in $$\Delta ADB, \angle DAB=\angle 45^o$$
$$\Rightarrow AD=DB$$(opposite sides are equal)
$$\Rightarrow AB^2+BC^2+2BD\times BC=AC^2$$
$$\Rightarrow AB^2+BC^2+2AD\times BC=AC^2$$
$$\Rightarrow AB^2+BC^2+4\left[\dfrac{1}{2}AD\times BC\right]=AC^2$$
$$\Rightarrow AB^2+BC^2+4(\Delta ABC)=AC^2$$
$$\therefore$$ Hence proved.
Show that in a right-angled triangle, the hypotenuse is the longest side.
Let ABC is a right angle triangle AC is hypotenuse
to show i) AC > AB
ii) AC > BC
$$\Delta ABC$$
$$\angle ABC=90^{\circ}$$
$$\angle ABC+\angle BAC+\angle ACB=180^{\circ}$$
$$\angle BAC$$ & $$\angle ACB$$ are acute angle.
$$\angle BAC < 90^{\circ}$$ & $$\angle ACB < 90^{\circ}$$
$$\angle BAC < \angle ABC$$ & $$\angle ACB < ABC$$
$$\therefore BC < AC $$ & $$ABC < AC$$ proved
Find $$x^{o}$$
Classify the above angle as right, straight, acute, obtuse or reflex.
In the given figure, check whether the angle measures $$30^{\circ}$$. Look at the same figure through a magnifying glass. Does the angle become larger? Does the size of the angle charge?
Step 1: To check
(1) Whether the angle measures $$30^o$$ or not?
(2) Whether the angle remains when we look it through a magnifying glass?
Step 2: Explanation
After measuring the angle with a protractor, we observe that the angle is $$30^\circ$$. [Image 1]
After using a magnifying glass, the angle doesn't change, the rays would change their length, hence the size of rays increases. When looked through a magnifying glass, the angle is again measured $$30^\circ$$. [Image 2]
Therefore, the angle remains the same in both cases.
Measure approximately and classify each angle .
| Angle | Measure | Type |
| $$\angle AOB$$ | ||
| $$\angle AOC$$ | ||
| $$\angle BOC$$ | ||
| $$\angle DOC$$ | ||
| $$\angle DOA$$ | ||
| $$\angle DOB$$ | - |
Find the measure of the angle shown in the each figure. (First estimate with your eyes and then find the actual measure with a protractor).
Which angle has a large measure? First estimate and then measure.
Measure of Angle $$A=$$
Measure of Angle $$B=$$
Write number of edges, faces and vertices of cube, cuboid, cone, cylinder, sphere, triangular pyramid, rectangular and prism.
$$PQR$$ is a triangle in which $$PQ = PR$$ and $$S$$ is any point on the side $$PQ.$$ Through $$S$$, a line is drawn parallel to $$QR$$ and intersecting $$PR$$ at $$T$$. proves that $$PS = PT$$.
$$PQ=PR\Rightarrow \angle R=\angle Q$$
Now, $$ST \parallel QR$$.
$$\Rightarrow \angle PST=\angle PQR$$ and $$\angle PTS=\angle PRQ$$ [$$\because $$ Corresponding angles are equal]
$$\Rightarrow \angle PST=\angle Q$$ and $$\angle PTS=\angle R$$
$$\Rightarrow \angle PST=\angle PTS$$
$$\Rightarrow PT=PS$$
Explain the following terms:
Scalene triangle
Triangle with different lengths of sides is called as scalene triangle.(shown in figure)
In Fig.the sides, BA and CA have been produced such that $$BA = AD$$ and $$CA = AE$$. Prove that segment $$DE \parallel BC$$.
If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.
$$\textbf{Step-1: Draw the appropriate diagram & label it.}$$
$$\text{Let}$$ $$ \angle A< \angle B +\angle C. $$ $$\text{Then,}$$ $$\text{Adding}$$ $$\angle A$$ $$\text{on both sides, we get,}$$
$$ 2\angle A < \angle A + \angle B +\angle C \Rightarrow 2\angle A < 180^{\circ} \Rightarrow \angle A < 90^{\circ}$$
$$[\textbf{ sum of three angles of a triangle =}$$ $$\boldsymbol{180^{\circ}}]$$
$$\text{Similarly, we obtain}$$ $$ \angle B < 90^{\circ}$$ $$\text{and}$$ $$ \angle C< 90^{\circ}$$
$$\Rightarrow$$ $$\text{Each angle of the triangle is less than 90°}$$
$$\textbf{Hence, it is an acute-angled triangle.}$$
Fill in the blanks to make the following statements true:
A triangles cannot have more then ______ obtuse angles.
Using protractor, draw a right angle. Bisect it to get an angle of measure $$45^{o}$$.
In Fig., there is a triangle. The measures of some angles have been indicated. State whether triangle is acute, right or obtuse.
Explain the following terms:
Obtuse triangle
Fill in the blanks with the correct word/symbol to make it a true statement:
A triangle whose one angle is more than $$90^{o}$$ is known as ...........
Triangle having one angle more than 90 degrees is obtuse angled or obtuse triangle.
Construct an angle of $$ 60^{\circ}$$ with the help of compasses and bisect it by paper folding.
Draw a ray $$OA$$.
With convenient radius and centre $$O$$, draw an arc cutting the ray $$OA$$ at $$P$$.
With the same radius and centre at $$P$$, draw another arc cutting the previous arc at $$Q$$.
Draw $$OQ$$ and extend it to $$B$$.
$$∠AOB$$ is the required angle of $$60^0$$.
The diagram is shown in the above image.
We cut the part of paper as sector $$OPQ$$.
Now, fold the part of paper such that line segments $$OP$$ and $$OQ$$ get coincided.
Angle made at point $$O$$ is the required angle, which is half of angle $$∠AOB$$.
The diagram is shown in the above image.
Fill in the blanks with the correct word/symbol to make it a true statement:
A triangle whose all the angle are of measure less than $$90^{o}$$ is known as .........
Triangle having all angles less than 90 degrees is acute angled triangle or acute triangle.(Shown in fig.)
Explain the following terms:
Right triangle
Fill in the blanks with the correct word/symbol to make it a true statement:
A triangle whose one angle is a right angle is known as ..........
Triangle having one right angle(90 degree) is called as right angled triangle.
Fill in the blanks to make the following statements true:
________edges of a cube (or cuboid) meet at each of its vertices.
Explain the following terms:
Acute triangle
What is the measure of the angle in degrees between North and South-East?
In figure, write another name for $$\angle 1$$.
Shikha is rowing a boat due north-east. In which direction will she be rowing if she turns it through a complete angle?
What is the measure of the angle in degrees between North and West?
The measure of the angle in degrees between North and West is $$90^0$$.(Refer figure)
What is the measure of the angle in degrees between North and South?
The measure of the angle in degrees between North and South is $$180^0$$.(Refer figure)
Construct the following angle with the help of ruler and compasses only:$$135^{\circ}$$
Draw the line $$AB$$ and take the point $$O$$ at the middle of $$AB$$.
With a convenient radius and centre at $$O$$, draw an arc, which cuts $$AB$$ at $$P$$ and $$Q$$, respectively.
Draw an angle of $$90^0$$ on the ray $$OB$$ as $$∠BOC = 90^0$$, where ray $$OC$$ cuts the arc at $$R$$.
With $$Q$$ and $$R$$ as centres and radius more than half of $$QR$$, draw two arcs, which cuts each other at $$S$$.
Draw $$OS$$ and extend it to form the ray $$OD$$.
$$∠BOD$$ is required angle of $$135^0$$.
Using only a ruler, draw an acute angle, a right angle and an obtuse angle in your notebook and name them.
Explain the following term.
Scalene triangle.
In the following, the measures of three angles are given. State whether the angles can possibly be those of a triangle.
(i) $$63^o, 37^o, 80^o$$.
(ii) $$45^o, 61^o, 73^o$$(iii) $$59^o, 72^o, 61^o$$(vi) $$30^o, 20^o, 125^o$$
In figure, line segments AB and CD bisect each other at O. State true or false. $$\Delta AOC\cong \Delta DOB$$.
State the three pairs of matching parts, you have used to arrive at the answer.
In the given figure, $$m$$ and $$n$$ are two plane mirrors perpendicular to each other. Show that the incident ray $$CA$$ is parallel to the reflected ray $$BD$$
Write down the number of faces in a cuboid
A cuboid has $$6$$ faces
Is it true to say that AB$$=$$AC?
Write down the number of faces of each of the following figures:
Tetrahedron
Write down the number of faces in a Cube
A cube has $$6$$ faces namely $$ABFE, BFGC, GHDC, HEAD, DCBA$$ and $$HGFE$$
Write down the number of edges of a Cube
A cube has $$12$$ edges.
Namely $$AB, BC, CD, DA, EF, FG, GH, HE, AE, DH, BF, CG$$
Write down the number of edges of a Tetrahedron
Tetrahedron has six edges.
Namely $$OA, OB, OC, AB, AC$$ and $$BC$$
State the type of each of the following angles:
Write down the number of edges of a Triangular prism
A triangular prism has $$9$$ edges.
Namely $$AB, BC, CD, DA, EF, FC, FD, EA, BE$$
How many edges are there in a Cuboid
A cuboid had $$12$$ edges.
Namely, $$AB, BC, CD, DA, EF, FG, GH, HE, AE, DH, BF, CG$$
Write down the number of edges of a Rectangular Prism
The rectangular pyramid has $$8$$ edges.
Namely $$AB, AC, AD, AE, DE, EB, BC$$ and $$CD$$
How many edges are there in a Tetrahedron
A tetrahedron has $$6$$ edges.
Namely $$OA, OB, OC, AB, AC$$ and $$BC$$
Classify the angle whose magnitude is given below:
$$90^{o}$$
Measures each of the following angles with the help of a protractor and write the measures in degree:
How many degrees are there in
three right angles?
How many degrees are there in
two right angles?
Classify the angles whose magnitudes are given below:
four right angles?
Measures each of the following angles with the help of a protractor and write the measures in degree:
How many degrees are there in
one right angles?
Measures each of the following angles with the help of a protractor and write the measures in degree:
Measures each of the following angles with the help of a protractor and write the measures in degree:
Fill in the blanks:
A cuboid has...................rectangular faces, ........................edges and............vertices.
In the given figure, compare the line segment with the help of a divider and fill in the blanks by using the symbol >, = or <
CD ..... BD
Measures each of the following angles with the help of a protractor and write the measures in degree:
In the given figure, verify the following by measurement:
If $$AB=3cm,BC=1.5cm,AC=4.5cm$$
$$AB + BC = AC$$
In the given figure, verify the following by measurement:
$$AC - BC = AB$$
Which of the following are models for perpendicular lines?
(i) The adjacent edges of a postcard.
(ii) The lines of a railway track.
(iii) The line segment forming the letter 'L'.
(iv) The adjacent edges of your Math book.
(v) The line segment forming thee letter 'V'
In the given figure, verify by measurement that :
$$AC + BD = AD + BC$$
In the figure given below, Line $$l$$ is perpendicular to line $$m.$$
Does $$\overset{\leftrightarrow }{PE}$$ disect segment $$\overline{BH}$$?
In the given figure, measure the lengths of the sides of the triangle ABC and verify :
If $$AB=2.5cm,BC=3.8cm$$ and $$AC=4cm,$$
$$AC + AB > BC$$
In the figure below, line $$l$$ is perpendicular to line $$m.$$
Is CE = EG?
What is the shape of a sweet laddu ?
Match the following:
Using the information given, name the right angles in each part of the figure.
What is common in the given figures (i) and (ii)?
Is figure (i) that of the triangle? If not, why?
In which of the given figures.
(a) A perpendicular bisector is shown?
(b) The bisector is shown
(c) The only bisector is shown
(d) The only perpendicular is shown?
Write the name of
(a) Vertices (b) edges (c) Faces of the prism shown in figure
How many edges, faces and vertices are there in a sphere?
In the figure, the number of faces meeting at $$B$$ is _____.
In the figure, the number of faces meeting at $$B$$ is $$4$$
In the given figure, faces $$ ACB, EAB, DCB$$ and $$EDB$$ meet at $$B$$.
In a three-dimensional shape, diagonal is a line segment that joins two vertices that do not lie on the _____ face.
If the sum of the number of vertices and faces in a polyhedron is $$14$$, then the number of edges in the shape is ______.
The number of cubes in figure are _____.
A regular polyhedron is a solid made up of _____ faces.
Every triangle has at least ______ acute angle (s).
How many vertices does the following solid have?
a) Cone
Cone: It has 1 vertex. The structure of cone is shown below:
How many vertices does the following solid have?
Tetrahedron
Tetrahedron: It has 4 vertices. The structure of tetrahedron is shown below:
How many vertices does the following solid have?
Sphere
How many vertices does the following solid have?
Octagonal Pyramid
Octagonal Pyramid: It has 9 vertexes. The structure of the octagonal pyramid is shown below:
How many faces does the following solid have?
Octahedron
Octahedron: It has $$8$$ faces. The structure of octahedron is shown below:
How many vertices does the following solid have?
Cylinder
It has zero vertices. The structure of cylinder is shown below:
How many faces does the following solid have?
Tetrahedron
Faces of the given solid are given below:
Tetrahedron : It has $$4$$ faces. The structure of tetrahedron is shown below:
Tetrahedron : It has $$4$$ faces. The structure of tetrahedron is shown below:
How many faces does the following solid have?
Hexahedron
Hexahedron : It has 6 faces. The structure of hexahedron is shown below:
How many faces does the following solid have?
Octagonal Pyramid
Octagonal Pyramid : It has 9 faces. The structure of octagonal pyramid is shown below:
How many edges does the following solid have?
Cone
Cone: It has 1 edge. The structure of cone is shown below:
How many edges do the following solid have?
Sphere
Sphere: It has no edge. The structure of sphere is shown below:
How many edges do the following solid have?
Cylinder
Cylinder: It has 2 edge. The structure of cylinder is shown below:
Find the number of faces in the given shapes:
A polyhedron has 60 edges and 40 vertices. Find the number of its faces.
A polyhedron has 20 faces and 12 vertices. Find the edges of the polyhedron.
How many edges do the following solid has?
Kaleidoscope
$$Ref$$ Image
The structure of the kaleidoscope is shown above. It has $$9 $$ edges.
Write the number of faces, vertices, and edges in
(i) an octagonal prism
(ii) decagonal pyramid.
The number of faces, vertices, and edges in the following figures are listed below:
No. of faces | No. of vertices | No. of edges | |
(i) an octagonal prism | 10 | 16 | 24 |
(ii) decagonal pyramid | 11 | 11 | 20 |
Which of the following can be the sides of a right angled triangle?
$$4cm, 5cm, 7cm$$
State If the given figure shows two mutually perpendicular lines.
State If the given figure shows two mutually perpendicular lines.
Using the given figure, answer the following:
Is the line $$q$$ perpendicular to the line $$m$$?
Name three objects from your surroundings that contain perpendicular edges.
Give two examples from your surroundings for the following:
perpendicular lines.
Using the given figure, answer the following:
Name the pairs of mutually perpendicular lines
State, which of the lines/line-segments are perpendicular to the line $$PQ$$:
Check your text book:
How many pairs of its edges are parallel to each other?
Check your text book:
How many pairs of its edges are perpendicular to each other?
State If the given figure shows two mutually perpendicular lines.
The figure, given, shows shadow of some 3D object, when seen under the lamp of an overhead projector. Name the object.
The following figure is considered to be the image of a 3D object when seen under the lamp of an overhead projector
Classify the following triangles according to angle:
Classify the following triangles according to angle:
State the number of faces , number of vertices and number of edges of :
a pentagonal pyramid
a pentagonal pyramid
Number of faces $$= 6$$
Number of vertices $$= 6$$
Number of edges $$= 10$$
Number of faces $$= 6$$
Number of vertices $$= 6$$
Number of edges $$= 10$$
What is the least number of planes that can enclose a solid? What is the name of the solid?
The figure, given, shows shadow of some 3D object, when seen under the lamp of an overhead projector. Name the object.
The following figure is considered to be the image of a 3D object when seen under the lamp of an overhead projector
State the number of faces , number of vertices and number of edges of :
A hexagonal prism
A hexagonal prism
Number of faces $$= 8$$
Number of vertices $$= 12 $$
Number of edges $$= 18$$
Number of faces $$= 8$$
Number of vertices $$= 12 $$
Number of edges $$= 18$$
If a polyhedron has 10 vertices and 7 faces , fins the number of edges in it.
The figure, given, shows shadow of some 3D object, when seen under the lamp of an overhead projector. Name the object.
The following figure is considered to be the image of a 3D object when seen under the lamp of an overhead projector.
Can a polyhedron have 8 faces, 26 edges and 16 vertices?
Is the following a model for perpendicular lines:
The line segments forming the letter $$'L'$$.
Which of the following are models for perpendicular lines :
The letter $$V$$.
Match the following :
Give two new examples of each shape.
Which of the following are models for perpendicular lines :
The adjacent angles of a table top.
The adjacent edge of a table top are perpendicular to each other.
Which of the following are models for perpendicular lines :
The line of a railway track.
Fill in the blanks with acute, right and straight :
When the sum of the measure of two angles is that of a right angle, then each one of them is _________.
What is the name given to the longest side of a right-angled triangle
Hypotenuse
Show that in a right angled triangle, the hypotenuse is the longest side.
Give a definition for each of the following terms. Are there other terms that need to be defined first ? What are they, and how might you define them ?
Perpendicular lines
Perpendicular lines:
AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.
Data: $$AD$$ and $$BC$$ are equal perpendiculars to a line segment $$AB$$.
To Prove:
$$CD$$ Bisects $$AB$$.
Proof: In $$\Delta CBO$$ and $$\Delta DAO$$,
$$BC = AD$$
$$\angle CBO = \angle DAO = 90^0$$
$$\angle BOC = \angle AOD$$ (Vertically opposite angles)
$$SAS$$ postulate,
$$\therefore \ \ \Delta CBO = \Delta DAO$$
$$ \therefore \ \ OA = OB$$
$$\therefore \ \ CD$$ bisects $$AB$$ at $$'O'$$.
To Prove:
$$CD$$ Bisects $$AB$$.
Proof: In $$\Delta CBO$$ and $$\Delta DAO$$,
$$BC = AD$$
$$\angle CBO = \angle DAO = 90^0$$
$$\angle BOC = \angle AOD$$ (Vertically opposite angles)
$$SAS$$ postulate,
$$\therefore \ \ \Delta CBO = \Delta DAO$$
$$ \therefore \ \ OA = OB$$
$$\therefore \ \ CD$$ bisects $$AB$$ at $$'O'$$.
Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
In right triangle the make the statement true
In right triangle the hypotenuse is the _____ side
Find foot of the $$\perp$$ from the point (2, 3) on the line $$3x-y+4=0$$
Use a graph paper for this question. Plot the points $$A(0,4)$$ and $$B(-4,0)$$. These two points are vertices of a figure which is symmetrical about the x-axis and y-axis. Complete the figure on the graph.
i) Give the geometrical name of the figure formed.
ii) Find the area of the figure.
Find measure of $$\\ \angle ABO$$ where O is the origin.
Points are $$A(0,4) $$ and $$B(-4,0)$$
By joining $$A$$ & $$B$$,
(i) Geometrical name of the figure formed$$=\triangle (AOB)=Right\,angled\,triangle$$.
(ii) Area of right angled triangle$$=\cfrac{1}{2}(4)\times (4)=8\,sq\,units$$.
Measure of $$\angle ABO=\angle BAO=90^{\circ}$$
as $$\angle AOB=90^{\circ}$$
So, $$\angle ABO=\cfrac{90}{2}=45^{\circ}$$
Here $$\triangle$$ABC is a right angle triangle with right angle at C. Let BC =a , Ca = b, AB = c. Let P be the length of the perpendicular from C on AB. Prove that (i) Pc = ab.
In $$\triangle ACD,$$
$$AD^{2}+CD^{2}=AC^{2}$$ (Using Pythagoras theorem)
$$\Rightarrow AD^{2}=b^{2}-p^{2}$$
$$\Rightarrow AD=\sqrt{b^{2}-p^{2}}$$--------------------(1)
In $$\triangle BCD,$$
$$BD^{2}+CD^{2}=BC^{2}$$ (Using Pythagoras theorem)
$$\Rightarrow BD^{2}=a^{2}-p^{2}$$
$$\Rightarrow BD=\sqrt{a^{2}-p^{2}}$$-------------------(2)
In $$\triangle ABC$$,
$$AC^{2}+BC^{2}=AB^{2}$$
$$b^{2}+a^{2}=c^{2}$$-------------------(3)
Now,
$$AD+BD=AB=C$$
$$\Rightarrow \sqrt{b^{2}-p^{2}}+\sqrt{a^{2}-p^{2}}=c$$
Squaring we get,
$$\Rightarrow b^{2}-p^{2}+a^{2}-p^{2}+2\sqrt{b^{2}-p^{2}}\sqrt{a^{2}-p^{2}}=c^{2}$$
$$\Rightarrow a^{2}+b^{2}-2p^{2}+2\sqrt{b^{2}-p^{2}}\sqrt{a^{2}-p^{2}}=a^{2}+b^{2}$$ $$[\because c^{2}=a^{2}+b^{2}]$$
$$\Rightarrow p^{2}=\sqrt{b^{2}-p^{2}}\sqrt{a^{2}-p^{2}}$$
Squaring,
$$\Rightarrow p^{4}=a^{2}b^{2}+p^{4}-p^{2}(a^{2}+b^{2})$$
$$\Rightarrow p^{2}(a^{2}+b^{2})=a^{2}b^{2}$$
$$\Rightarrow p^{2}c^{2}=a^{2}b^{2}$$ $$(\because c^{2}=a^{2}+b^{2})$$
$$\Rightarrow pc=ab$$ (Taking the square root).
Name the types of following triangles:
$$\triangle ABC$$ with $$AB=8.7\ cm,AC=7\ cm$$ and $$BC=6\ cm$$
Draw a rough figure and label suitably in each of the following cases:-
a)point R lies on $$\Delta ABC$$.
b) Line XY and line PQ interest at M.
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Determine whether the triangle having sides $$(a-1)\ cm$$, $$2\sqrt{a}\ cm$$ and $$(a+1)\ cm$$ is a right angled triangle or not.
Find $$x$$ if $$\triangle ABC$$ is right angled at $$A$$, where $$A\equiv(4, 2, 3), B\equiv(3, 1, 8), C\equiv(x, -1, 2)$$.
Prove that DBCDBC is right angle
Verify that points $$P(-2,2), Q(2,2)$$ and $$R(2,7)$$ are vertices of right angled triangle.
Prove that $$\overline { a } = ( 4 , - 3,1 ) , \overline { b } = ( 2 , - 4,5 )$$ and $$\overline { c } = ( 1 , - 1,0 )$$ are vertices of a right angle.
If $$A\left(h,k\right),B\left(1,1\right),C\left(2,1\right)$$ are the vertices of a right angled triangle with $$AC$$ as its hypotenuse and the area of the triangles is $$1\ sq$$, unit, then find the value $$(s)$$ of $$k$$.
Let $$\vec { a } , \vec { b } , \vec { c }$$ be position vectors of three vertices of triangle $$A B C ,$$ find the area of triangle $$ABC.$$
$$\cos\theta =\dfrac { \overrightarrow { AD } }{ AB } =\dfrac { \left| \overrightarrow { AD } \right| }{ \left| \overrightarrow { b } -\overrightarrow { a } \right| } $$
$$\Rightarrow \left| \overrightarrow { AD } \right| =$$ height $$=\left| \overrightarrow { b } -\overrightarrow { a } \right| \cos\theta $$
and $$\cos\theta =\dfrac { \left( \overrightarrow { c } -\overrightarrow { b } \right) .\left( \overrightarrow { a } -\overrightarrow { b } \right) }{ \left| \overrightarrow { c } -\overrightarrow { b } \right| \left| \overrightarrow { a } -\overrightarrow { b } \right| } $$ (Using $$\cos\theta =\dfrac { \overrightarrow { a } .\overrightarrow { b } }{ \left| \overrightarrow { a } \right| \left| \overrightarrow { b } \right| } $$)
$$\therefore$$ height $$=\left| \overrightarrow { b } -\overrightarrow { a } \right| \left( \dfrac { \left( \overrightarrow { c } -\overrightarrow { b } \right) .\left( \overrightarrow { a } -\overrightarrow { b } \right) }{ \left| \overrightarrow { c } -\overrightarrow { b } \right| \left| \overrightarrow { a } -\overrightarrow { b } \right| } \right) =\dfrac { \left( \overrightarrow { c } -\overrightarrow { b } \right) .\left( \overrightarrow { a } -\overrightarrow { b } \right) }{ \left| \overrightarrow { c } -\overrightarrow { b } \right| } $$
$$\therefore$$ area of triangle $$=\dfrac { 1 }{ 2 } \times $$ base height
$$=\dfrac { 1 }{ 2 } \left( \overrightarrow { c } -\overrightarrow { b } \right) \left\{ \dfrac { \left( \overrightarrow { c } -\overrightarrow { b } \right) .\left( \overrightarrow { a } -\overrightarrow { b } \right) }{ \left| \overrightarrow { c } -\overrightarrow { b } \right| } \right\} $$
Fill in the blank.In quadrilateral ROPE the pairs of adjacent angles are ___.
Construct the following angle with the help of ruler and compasses only:$$45^{\circ}$$
To construct an angle of $$45^0$$, construct an angle of $$90^0$$ and bisect it.
$$\Rightarrow$$Procedure to draw $$\angle90^{\circ}$$
Draw a ray $$OA$$.
With a convenient radius and centre at $$O$$, draw an arc cutting the ray $$OA$$ at $$P$$.
With the same radius and centre at $$P$$, draw another arc, which cuts the first arc at $$Q$$.
With the same radius and centre at $$Q$$, draw another arc, which cuts the first arc at $$R$$.
With $$Q$$ and $$R$$ as centres and radius more than half of $$QR$$, which cuts each other at $$S$$.
Draw $$OS$$ and extend it to $$B$$ from the ray $$OB$$.
$$∠AOB$$ is required angle of $$90^0$$.
After constructing the $$∠AOB = 90^0$$, rays $$OA$$ and $$OB$$ intersect the arc at points $$P$$ and $$T$$ as shown in figure.
With $$P$$ and $$T$$ as centers and radius more than half of $$PT$$, draw two arcs, which cut each other at $$X$$.
Draw $$OX$$ and extend it to $$C$$ to form the ray $$OC$$.
$$∠AOC$$ is the required angle of $$45^0$$.
Construct the following angle with the help of ruler and compasses only:$$90^{\circ}$$
Draw a ray $$OA$$.
With a convenient radius and centre at $$O$$, draw an arc cutting the ray $$OA$$ at $$P$$.
With the same radius and centre at $$P$$, draw another arc, which cuts the first arc at $$Q$$.
With the same radius and centre at $$Q$$, draw another arc, which cuts the first arc at $$R$$.
With $$Q$$ and $$R$$ as centres and radius more than half of $$QR$$, which cuts each other at $$S$$.
Draw $$OS$$ and extend it to $$B$$ from the ray $$OB$$.
$$∠AOB$$ is required angle of $$90^0$$.
Line-segments AB and CD bisect each other at O. AC and BD are joined forming triangles AOC and BOD. State the three equality relations between the parts of the two triangles, that are given or otherwise known. Are the two triangles congruent? State in symbolic form. Which congruence condition do you use?
Let, this be figure of given condition.
In figure, line segments AB and CD bisect each other at O.
So, $$O$$ is midpoint of $$CD$$ and $$AB$$
Therefore,
$$OC=OD$$
$$OB=OA$$
and $$\angle AOC\cong\angle BOD\cdots\cdots[\text{vertically opposite angles}]$$
Hence, by $$SAS$$ test we can say that
$$\Delta AOC\cong \Delta BOD$$
So, yes two triangles are congruent.
Congruence condition used is $$SAS$$
Above written two pairs of equal sides and one pair of equal angles are three equality relations of matching parts used to arrive at answer
AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B (See Fig. 12.26). Show that the line PQ is perpendicular bisector of AB.
Using a protractor, draw a right angle and bisect it to get an angle of measure $$ 45^{\circ}.$$
We are asked to construct an angle of 90° using protractor and bisect it.
We will follow the following algorithm to construct this figure.
Steps of Construction
STEP1: Using the protractor, draw a right angle and label it as angle $$ABC$$.
STEP2: With centre as $$B$$, draw an arc of any radius, intersecting the ray $$BA$$ at $$P$$ and the ray $$BC$$ at $$Q$$.
STEP3: With centre as $$P$$, and taking radius greater than half of $$PQ$$, draw an arc inside the angle $$ABC$$.
STEP4: With centre as $$Q$$, and taking the same radius, draw another arc, intersecting the previous arc at $$R$$.
STEP5: Draw the ray $$BR$$.
In figure, line CA$$\perp$$AB$$||$$ line CR and line PR $$||$$ line BD. Find $$\angle x, \angle y$$ and $$\angle z$$.
Construct the following angle with the help of ruler and compasses only: $$30^{\circ}$$
Draw a ray $$OA$$.
With a convenient radius and centre at $$O$$, draw an arc, which cuts $$OA$$ at $$P$$.
With the same radius and centre at $$P$$, draw an arc cutting the previous arc at $$Q$$.
Taking $$P$$ and $$Q$$ as centres and radius more than half of $$PQ$$, draw two arcs, which cuts each other at $$R$$.
Draw $$OR$$ and extend it to $$B$$.
$$∠AOB$$ is the required angle of $$30^0$$.
Show that the points (0,7,10), (-1, 6, 6) and (-4, 9,6) form a right angled triangle.
Construct the following angles with the help of a protractor: $$67^{\circ},38^{\circ},110^{\circ},179^{\circ},98^{\circ},84^{\circ}$$
(i) $$67^°$$
We draw a ray $$OA$$.
We place the protractor on $$OA$$ such that its center coincides with the point $$O$$ and the diameter of the protractor coincides with $$OA$$.
We mark a point $$B$$ against the mark of $$67^°$$ on the protractor.
We remove the protractor and draw $$OB. ∠AOB$$ is the required angle of $$67^°$$.
Similarly, we draw the angles $$38^°,110^°,179^°,98^°,84^°$$ as shown in the above image.
Using a protractor, draw an angle of measure $$ 72^{\circ}.$$ With this angle as given, draw angles of measure $$ 36^{\circ}$$ and $$ 54^{\circ}.$$
Draw a ray $$OA$$.
With the help of a protractor, draw an angle $$∠AOB$$ of $$72^°$$.
With a convenient radius and centre at $$O$$, draw an arc cutting sides $$OA$$ and $$OB$$ at $$P$$ and $$Q$$, respectively.
With $$P$$ and $$Q$$ as centres and radius more than half of $$PQ$$, draw two arcs cutting each other at $$R$$.
Join $$O$$ and $$R$$ and extend it to $$X$$.
$$OR$$ intersects arc $$PQ$$ at $$C$$.
With $$C$$ and $$Q$$ as centres and radius more than half of $$CQ$$, draw two arcs cutting each other at $$T$$.
Join $$O$$ and $$T$$ and extend it to $$Y$$.
Now, $$OX$$ bisects $$∠AOB$$
Therefore, $$∠AOX = ∠BOX = \dfrac {72}{2}=36^°$$
Again, $$OY$$ bisects $$∠BOX$$
Therefore, $$∠XOY = ∠BOY = \dfrac {36}{2} = 18^°$$
Therefore, $$∠AOX$$ is the required angle of $$36^°$$ and $$∠AOY = ∠AOX + ∠XOY = 36^° + 18^° = 54^°$$
The diagram is shown the above image.
Therefore, $$∠AOY$$ is the required angle of $$54^°$$.
Class 6 Maths Extra Questions
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